Laws of Chemical Combination
The laws of chemical combination are fundamental principles that describe how elements combine to form compounds. These laws are essential for understanding chemical reactions and stoichiometry. Here are the five primary laws:
1. Law of Conservation of Mass
Statement: Proposed by Antoine Lavoisier in 1789, this law states that mass is neither created nor destroyed in a chemical reaction. The total mass of reactants equals the total mass of products.
Mathematical Representation: [ \text{Mass of reactants} = \text{Mass of products} ]
Example: When 5.2 g of calcium carbonate (CaCO₃) is heated, it decomposes to form 3.2 g of calcium oxide (CaO) and 1.99 g of carbon dioxide (CO₂). The total mass of products (3.2 g + 1.99 g = 5.19 g) is nearly equal to the mass of the reactant (5.2 g), considering experimental errors1.
Application in Stoichiometry: This law is crucial for balancing chemical equations. For example, in the reaction: [ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} ] The total mass of reactants (propane and oxygen) equals the total mass of products (carbon dioxide and water).
2. Law of Definite Proportions
Statement: Proposed by Joseph Proust in 1799, this law states that a given chemical compound always contains its component elements in a fixed ratio by mass, regardless of its source or method of preparation.
Mathematical Representation: [ \text{Mass ratio} = \frac{\text{Mass of element A}}{\text{Mass of element B}} ]
Example: Water (H₂O) always contains hydrogen and oxygen in a mass ratio of 1:8. This means that 9 g of water will always contain 1 g of hydrogen and 8 g of oxygen1.
Application in Chemical Analysis: This law is used to determine the purity of compounds. For instance, if a sample of water does not have a 1:8 mass ratio of hydrogen to oxygen, it indicates the presence of impurities.
3. Law of Multiple Proportions
Statement: Proposed by John Dalton in 1803, this law states that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.
Mathematical Representation: [ \frac{\text{Mass of element A in compound 1}}{\text{Mass of element A in compound 2}} = \text{Simple whole number ratio} ]
Example: Carbon forms two oxides with oxygen: carbon monoxide (CO) and carbon dioxide (CO₂). In CO, 12 g of carbon combines with 16 g of oxygen (ratio 12:16 or 3:4). In CO₂, 12 g of carbon combines with 32 g of oxygen (ratio 12:32 or 3:8). The ratio of oxygen masses (16:32) is a simple whole number ratio of 1:21.
Application in Molecular Formulas: This law helps in determining the empirical formulas of compounds. For example, the empirical formula of a compound can be derived by comparing the mass ratios of its elements.
4. Gay-Lussac’s Law of Gaseous Volumes
Statement: Proposed by Joseph Louis Gay-Lussac in 1808, this law states that when gases react together at constant temperature and pressure, the volumes of the reacting gases and the volumes of the products (if gaseous) are in simple whole number ratios.
Mathematical Representation: [ \frac{\text{Volume of gas A}}{\text{Volume of gas B}} = \text{Simple whole number ratio} ]
Example: When hydrogen gas reacts with oxygen gas to form water vapor, 2 volumes of hydrogen react with 1 volume of oxygen to produce 2 volumes of water vapor (2H₂ + O₂ → 2H₂O). The volume ratio of hydrogen to oxygen to water vapor is 2:1:21.
Application in Gas Reactions: This law is used to predict the volumes of gases involved in reactions. For example, in the synthesis of ammonia (NH₃) from nitrogen and hydrogen gases, the volume ratio is 1:3:2 (N₂ + 3H₂ → 2NH₃).
5. Avogadro’s Law
Statement: Proposed by Amedeo Avogadro in 1811, this law states that equal volumes of all gases, at the same temperature and pressure, contain an equal number of molecules.
Mathematical Representation: [ V \propto n ] where ( V ) is the volume and ( n ) is the number of moles.
Example: 1 liter of hydrogen gas and 1 liter of nitrogen gas, at the same temperature and pressure, contain the same number of molecules. This law helps in understanding the molecular composition of gases and is fundamental to the concept of the mole1.
Application in Molar Volume: Avogadro’s law is used to determine the molar volume of gases. At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters.
These laws form the basis of stoichiometry.
Application of “Laws of Chemical Combination” to solve numerical problems:
Applying the laws of chemical combination to solve numerical problems involves understanding the principles behind each law and using them to set up and solve equations. Here are some examples to illustrate how you can apply these laws:
1. Law of Conservation of Mass
Problem: When 10 g of calcium carbonate (CaCO₃) is heated, it decomposes into calcium oxide (CaO) and carbon dioxide (CO₂). If 5.6 g of CaO is formed, what mass of CO₂ is produced?
Solution:
2. Law of Definite Proportions
Problem: A sample of pure water is found to contain 2 g of hydrogen and 16 g of oxygen. Verify the law of definite proportions.
Solution:
3. Law of Multiple Proportions
Problem: Carbon forms two oxides: CO and CO₂. In CO, 12 g of carbon combines with 16 g of oxygen. In CO₂, 12 g of carbon combines with 32 g of oxygen. Show that these data obey the law of multiple proportions.
Solution:
4. Gay-Lussac’s Law of Gaseous Volumes
Problem: When 2 liters of hydrogen gas react with 1 liter of oxygen gas, how many liters of water vapor are produced?
Solution:
5. Avogadro’s Law
Problem: If 5 liters of nitrogen gas (N₂) at STP contain (1.34 \times 10^{23}) molecules, how many molecules are there in 10 liters of nitrogen gas at the same conditions?
Solution:
By understanding and applying these laws, you can solve a wide range of numerical problems in chemistry.
Real Life Application of Laws of Chemical Combination:
The laws of chemical combination have numerous real-life applications across various fields. Here are some examples:
1. Law of Conservation of Mass
Chemical Manufacturing: In industrial chemical processes, the law of conservation of mass is crucial for designing reactors and ensuring that the input and output masses are balanced. This helps in optimizing the yield and efficiency of chemical reactions.
Environmental Science: This law is used to track pollutants in the environment. For instance, in waste management, the total mass of waste before and after treatment must be accounted for to ensure no mass is lost or unaccounted for.
Cooking: When baking a cake, the total mass of the ingredients (flour, sugar, eggs, etc.) equals the mass of the final baked cake plus any gases released during baking (like CO₂ from baking powder).
2. Law of Definite Proportions
Pharmaceuticals: The law ensures that drugs are manufactured with precise chemical compositions. For example, a specific medication must contain the exact ratio of active ingredients to be effective and safe.
Food Industry: This law is applied in the formulation of food products to maintain consistent taste and quality. For example, table salt (NaCl) always contains sodium and chlorine in a 1:1 ratio by moles.
3. Law of Multiple Proportions
Chemical Synthesis: This law helps chemists understand and predict the formation of different compounds from the same elements. For example, nitrogen and oxygen can form NO, NO₂, N₂O, etc., each with different properties and uses.
Material Science: In the development of new materials, understanding the multiple proportions of elements can lead to the creation of compounds with desired properties, such as different oxides of iron used in various applications.
4. Gay-Lussac’s Law of Gaseous Volumes
Industrial Gas Production: This law is used in the production and storage of industrial gases. For example, in the synthesis of ammonia (NH₃) from nitrogen and hydrogen gases, the volume ratios help in designing reactors and optimizing conditions.
Respiratory Therapy: In medical applications, this law helps in understanding the volumes of gases involved in respiratory treatments, ensuring patients receive the correct mixture of oxygen and other gases.
5. Avogadro’s Law
Chemical Reactions: Avogadro’s law is fundamental in stoichiometry for calculating the volumes of gases involved in reactions. For example, in the Haber process for ammonia production, knowing the volume of nitrogen and hydrogen gases helps in optimizing the reaction conditions.
Balloon Inflation: When inflating balloons with gases like helium, Avogadro’s law helps in determining the amount of gas needed to achieve a desired volume at a given temperature and pressure.
Breathing: The law explains why the volume of air we inhale and exhale remains consistent under the same conditions, which is crucial for understanding lung function and respiratory health.
Industrial Application of Laws of Chemical Combination:
The laws of chemical combination play a crucial role in various industrial processes. Here’s how each law impacts chemical reactions in industry:
1. Law of Conservation of Mass
Industrial Chemical Reactions: This law ensures that all mass inputs are accounted for in the outputs. In industries like pharmaceuticals, petrochemicals, and food processing, precise measurements of reactants and products are essential to maintain efficiency and safety. For example, in the production of sulfuric acid (H₂SO₄) via the Contact Process, the total mass of sulfur dioxide (SO₂), oxygen (O₂), and water (H₂O) must equal the mass of sulfuric acid produced.
Waste Management: Industries use this law to ensure that waste products are minimized and properly accounted for. This is crucial for environmental compliance and sustainability.
2. Law of Definite Proportions
Quality Control: This law is vital in ensuring that compounds are produced with consistent composition. For example, in the pharmaceutical industry, drugs must contain precise amounts of active ingredients to be effective and safe. Any deviation from the fixed ratio can lead to ineffective or harmful products.
Material Consistency: In the production of alloys, such as steel, maintaining a consistent ratio of iron to carbon is essential for achieving the desired properties of the final product.
3. Law of Multiple Proportions
Product Development: This law helps in the development of new compounds with desired properties. For instance, different oxides of nitrogen (NO, NO₂) are used in various applications, from fertilizers to explosives. Understanding the ratios in which elements combine allows chemists to tailor compounds for specific uses.
Chemical Formulation: In industries like cosmetics and food, this law helps in formulating products with precise ingredient ratios to achieve the desired effects and maintain safety standards.
4. Gay-Lussac’s Law of Gaseous Volumes
Gas Production and Storage: This law is used in industries that produce and store gases, such as the production of ammonia (NH₃) in the Haber process. Knowing the volume ratios of nitrogen and hydrogen gases helps in designing efficient reactors and storage systems.
Combustion Processes: In industries like power generation and automotive manufacturing, this law helps in optimizing combustion reactions. For example, the correct volume ratio of air to fuel is crucial for efficient combustion in internal combustion engines.
5. Avogadro’s Law
Gas Measurement: This law is fundamental in industries that deal with gases. For example, in the production of industrial gases like oxygen and nitrogen, Avogadro’s law helps in determining the volumes of gases required and produced.
Chemical Engineering: In processes like the synthesis of polymers, understanding the volume and number of molecules involved helps in scaling up reactions from the laboratory to industrial scale.
Real-Life Industrial Examples
These laws are fundamental to the efficiency, safety, and sustainability of industrial chemical processes. They ensure that reactions are predictable, controllable, and scalable, which is essential for industrial applications.
Comments:
Questions based on the laws of chemical combination in JEE Advanced typically test a student’s understanding of fundamental principles and their ability to apply these principles to solve complex problems. Here are some types of questions and examples from past JEE Advanced papers:
Types of Questions
Examples from Past JEE Advanced Papers
Example 1: Law of Conservation of Mass
Question: When 10 g of calcium carbonate (CaCO₃) is heated, it decomposes into calcium oxide (CaO) and carbon dioxide (CO₂). If 5.6 g of CaO is formed, what mass of CO₂ is produced?
Solution:
Example 2: Law of Definite Proportions
Question: A sample of pure water is found to contain 2 g of hydrogen and 16 g of oxygen. Verify the law of definite proportions.
Solution:
Example 3: Law of Multiple Proportions
Question: Carbon forms two oxides with oxygen: CO and CO₂. In CO, 12 g of carbon combines with 16 g of oxygen. In CO₂, 12 g of carbon combines with 32 g of oxygen. Show that these data obey the law of multiple proportions.
Solution:
Example 4: Gay-Lussac’s Law of Gaseous Volumes
Question: When 2 liters of hydrogen gas react with 1 liter of oxygen gas, how many liters of water vapor are produced?
Solution:
Example 5: Avogadro’s Law
Question: If 5 liters of nitrogen gas (N₂) at STP contain (1.34 \times 10^{23}) molecules, how many molecules are there in 10 liters of nitrogen gas at the same conditions?
Solution:
These examples illustrate the types of questions you might encounter in JEE Advanced based on the laws of chemical combination. Practicing such problems will help you understand the application of these laws and prepare effectively for the exam.
Q 1. Which of the following pair of substances contain element and compound within a pair ?(A) \(\mathrm{O}_{2}, \mathrm{CH}_{4}\)(B) \(\mathrm{H}_{2}, \mathrm{O}_{2}\)(C) \(\mathrm{N}_{2}, \mathrm{CO}_{2}\)(D) \(\mathrm{Na}, \mathrm{CO}\); |
(d) All of these; |
(c) \(\mathrm{C}\) and \(\mathrm{D}\); |
(b) B only; |
(a) A, C, D; |
Q 2. Many countries use Fahrenheit scale for expressing temperature of atmosphere If temperature in any such country is measured \(41^{\circ} \mathrm{F}\) then what is its value in celcius scale and would you expect hot or cold atmosphere in that country?; |
(d) \(41^{\circ} \mathrm{C}\), hot; |
(c) \(5^{\circ} \mathrm{C}\), cold; |
(b) \(25^{\circ} \mathrm{C}\), normal; |
(a) \(15^{\circ} \mathrm{C}\), cold; |
Q 3. 1 litre of a hydrocarbon weighs as much as one litre of \(\mathrm{CO}_{2}\) at same \(\mathrm{P}\) and \(\mathrm{T}\). Then the molecular formula of the hydrocarbon is:; |
(D) \(\mathrm{C}_{3} \mathrm{H}_{6}\); |
(C) \(\mathrm{C}_{2} \mathrm{H}_{4}\); |
(B) \(\mathrm{C}_{2} \mathrm{H}_{6}\); |
(A) \(\mathrm{C}_{3} \mathrm{H}_{8}\); |
Q 4. The number of significant figures for the three numbers \(161 \mathrm{~cm}, 0 161 \mathrm{~cm}, 0 0161 \mathrm{~cm}\) are; |
(d) 3,3 and 3 respectively; |
(c) 3,3 and 4 respectively; |
(b) 3,4 and 4 respectively; |
(a) 3,4 and 5 respectively; |
Q 5. Irrespective of the source, pure sample, of water always yields \(88 89 \%\) mass of oxygen and \(11 11 \%\) mass of hydrogen This is explained by the law of; |
(d) constant volume; |
(c) constant composition; |
(b) multiple proportions; |
(a) conservation of mass; |
Q 6. The average atomic mass of neon based on following data is: \({ }^{20} \mathrm{Ne}\) \({ }^{21} \mathrm{Ne}\) Relative abundance \({ }^{22} \mathrm{Ne}\); |
(a) \(0.33 \mathrm{u}\); |
(b) \(20.187 \mathrm{u}\); |
(c) \(6.729 \mathrm{u}\); |
(d) \(18.058 \mathrm{u}\); |
Q 7. Which of the following statements indicates that law of multiple proportion is being followed; |
(d) At constant temperature and pressure \(200 \mathrm{~mL}\) of hydrogen will combine with \(100 \mathrm{~mL}\) oxygen to produce \(200 \mathrm{~mL}\) of water vapour.; |
(c) When magnesium burns in oxygen, the amount of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formed.; |
(b) Carbon forms two oxides namely \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\), where masses fo oxygen which combine with fixed mass of carbon are in the simple ration \(2: 1\).; |
(a) Sample of carbon dioxide taken from any source will always have carbon and oxygen in the ratio \(1: 2\).; |
Q 8. \(10 \mathrm{~g}\) of hydrogen and \(64 \mathrm{~g}\) of oxygen were filled in a steel vessel and exploded Amount of water produced in this reaction will be:; |
(a) \(3 \mathrm{~mol}\); |
(b) \(4 \mathrm{~mol}\); |
(c) \(1 \mathrm{~mol}\); |
(d) \(2 \mathrm{~mol}\); |
Q 9. A 5 molar solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is diluted from 1 litre to a volume of 10 litres, the normality of the solution will be :; |
(a) \(1 \mathrm{~N}\); |
(b) \(0.1 \mathrm{~N}\); |
(c) \(5 \mathrm{~N}\); |
(d) \(0.5 \mathrm{~N}\); |
Q 10. In a compound \(\mathrm{C}, \mathrm{H}\) and \(\mathrm{N}\) atoms are present in \(9: 1: 3 5\) by weight Molecular weight of compound is 108 Molecular formula of compound is; |
(a) \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{~N}_{2}\); |
(b) \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{~N}\); |
(c) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{~N}_{2}\); |
(d) \(\mathrm{C}_{9} \mathrm{H}_{12} \mathrm{~N}_{3}\).; |
Atomic and Molecular Masses
Understanding atomic and molecular masses is fundamental in chemistry, especially for solving problems related to stoichiometry, chemical reactions, and molecular structure. Here’s a detailed overview:
1. Atomic Mass
Definition: Atomic mass (also known as atomic weight) is the mass of an atom, typically expressed in atomic mass units (amu). One atomic mass unit is defined as one-twelfth the mass of a carbon-12 atom.
Atomic Mass Unit (amu):
Relative Atomic Mass:
Calculation Example: \([ \text{Relative Atomic Mass of Cl} = \frac{(75.77 \times 34.969) + (24.23 \times 36.966)}{100} \approx 35.5 \text{ amu} ]\)
2. Molecular Mass
Definition: Molecular mass (or molecular weight) is the sum of the atomic masses of all the atoms in a molecule. It is also expressed in atomic mass units (amu).
Calculation:
Examples:
3. Molar Mass
Definition: Molar mass is the mass of one mole of a substance (element or compound) expressed in grams per mole (g/mol). It is numerically equal to the molecular or atomic mass in amu.
Calculation:
Examples:
4. Applications in Stoichiometry
Balancing Chemical Equations:
Example: \([ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} ]\)
Using Molar Mass in Calculations:
Example:
5. Isotopes and Average Atomic Mass
Isotopes:
Average Atomic Mass:
6. Practical Applications
Chemical Reactions:
Pharmaceuticals:
Environmental Science:
Summary
Understanding atomic and molecular masses is essential for solving problems in stoichiometry, balancing chemical equations, and performing various calculations in chemistry
Additional Applications of Atomic and Molecular Mass
The concepts of atomic and molecular mass are used in various fields beyond basic chemistry. Here are some additional areas where these concepts are crucial:
Real-Life Examples
Industrial Usage
These examples illustrate the wide-ranging applications of atomic and molecular masses in various fields, highlighting their importance in both scientific research and industrial processes. If you have any specific questions or need further details, feel free to ask!
Q 1. Which of the following statements about a compound is incorrect?; |
(a) A molecule of a compound has atoms of different elements.; |
(b) A compound cannot be separated into its constituent elements by physical methods of separation.; |
(c) A compound retains the physical properties of its constituent elements.; |
(d) The ratio of atoms of different elements in a compound is fixed.; |
Q 2. A sample was weighted using two different balances The results were (i) \(3.929 \mathrm{~g}\) (ii) \(4.0 \mathrm{~g}\) How would the weight of the sample be reported?; |
(a) \(3.93 \mathrm{~g}\); |
(b) \(3 \mathrm{~g}\); |
(c) \(3.9 \mathrm{~g}\); |
(d) \(3.929 \mathrm{~g}\); |
Q 3. A complex of iron contains \(45.6 \%\) iron by mass. Find out the number of iron atoms in \(5.00 \mathrm{~g}\) of this complex number of iron atoms \(=x^{*} 10^{\wedge} 22\) find the \(x\); |
(A) 2.1; |
(B) 2.2; |
(C) 2.3; |
(D) 2.4; |
Q 4. Given \(\mathrm{P}=0 0030 \mathrm{~m}, \mathrm{Q}=2 40 \mathrm{~m}, \mathrm{R}=3000 \mathrm{~m}\), Significant figures in \(\mathrm{P}, \mathrm{Q}\) and \(\mathrm{R}\) are respectively; |
(a) \(2,2,1\); |
(b) \(2,3,4\); |
(c) \(4,2,1\); |
(d) \(4,2,3\); |
Q 5. The percentage of copper and oxygen in samples of \(\mathrm{CuO}\) obtained by different methods were found to be the same The illustrate the law of; |
(a) constant proportions; |
(b) conservation of mass; |
(c) multiple proportions; |
(d) reciprocal proportions; |
Q 6. What is the average atomic mass of bromine from the following data : (abundance is in \%) Isotope & Mass & Abundance \\ \({ }^{79} \mathrm{Br}\) & 78.9183361 & 50.69 \\ \({ }^{81} \mathrm{Br}\) & 80.916289 & 49.31; |
(a) 79.9; |
(b) 76.6; |
(c) 75.9; |
(d) 69.9; |
Q 7. The number of molecules in 8 96 litre of a gas at \(0^{\circ} \mathrm{C}\) and 1 atm pressure is approximately; |
(a) \(6.023 \times 10^{23}\); |
(b) \(12.04 \times 10^{23}\); |
(c) \(18.06 \times 10^{23}\); |
(d) \(24.08 \times 10^{22}\); |
Q 8. Which has the maximum number of molecules among the following?; |
(a) \(44 \mathrm{~g} \mathrm{CO}_{2}\); |
(b) \(48 \mathrm{~g} \mathrm{O}_{3}\); |
(c) \(8 \mathrm{~g} \mathrm{H}_{2}\); |
(d) \(64 \mathrm{~g} \mathrm{SO}_{2}\); |
Q 9. With increase of temperature, which of these changes?; |
(a) Molality; |
(b) Weight fraction of solute; |
(c) Molarity; |
(d) Mole fraction; |
Q 10. The empirical formula of an acid is \(\mathrm{CH}_{2} \mathrm{O}_{2}\), the probable molecular formula of acid may be :; |
(a) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{4}\); |
(b) \(\mathrm{CH}_{2} \mathrm{O}\); |
(c) \(\mathrm{CH}_{2} \mathrm{O}_{2}\); |
(d) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\); |
Mole Concept and Molar Masses
Understanding the mole concept and molar masses is crucial for solving problems in stoichiometry, chemical reactions, and quantitative analysis in chemistry. Here’s a detailed overview:
1. Mole Concept
Definition: The mole is a fundamental unit in chemistry that represents a specific number of particles (atoms, molecules, ions, etc.). One mole of any substance contains exactly \(( 6.02214076 \times 10^{23} )\) elementary entities, known as Avogadro’s number.
Avogadro’s Number: \([ N_A = 6.02214076 \times 10^{23} ]\)
Mole and Mass Relationship:
Calculating Moles: \([ \text{Number of moles} = \frac{\text{Mass of substance (g)}}{\text{Molar mass (g/mol)}} ]\)
Example:
2. Molar Mass
Definition: Molar mass is the mass of one mole of a substance (element or compound) expressed in grams per mole (g/mol). It is numerically equal to the molecular or atomic mass in atomic mass units (amu).
Calculation:
Examples:
3. Applications in Stoichiometry
Balancing Chemical Equations:
Example: \([ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} ]\)
Using Molar Mass in Calculations:
Example:
4. Concentration Terms
Molarity (M):
Example:
Molality (m):
Example:
5. Empirical and Molecular Formulas
Empirical Formula:
Example:
Molecular Formula:
Example:
Determining Molecular Formula:
Example:
6. Practical Applications
Chemical Reactions:
Pharmaceuticals:
Environmental Science:
Additional Applications of Mole Concept and Molar Mass
The concepts of mole and molar mass are used in various fields beyond basic chemistry. Here are some additional areas where these concepts are crucial:
Real-Life Examples
Q 1. Choose the correct combination & Element & Compound & Mixture \\; |
(a) & Ammonia & Sodium & Air \\; |
(b) & Water & Sugar & Aqueous sugar solution \\; |
(c) & Hydrogen & Oxygen & Water \\; |
(d) & Silver & Water & Air; |
Q 2. Two students performed the same experiment separately and each one of them recorded two readings of mass which are given below Correct reading of mass is \(3 0 \mathrm{~g}\) On the basis of given data, mark the correct option out of the following statements Students & \multicolumn{2}{c}{ Readings } \\ & (i) & (ii) \\ A & 3.01 & 2.99 \\ B & 3.05 & 2.95; |
(a) Results of both the students are neither accurate nor precise.; |
(b) Results of student \(\mathrm{A}\) are both precise and accurate.; |
(c) Results of student \(\mathrm{B}\) are neither precise nor accurate.; |
(d) Results of student B are both precise and accurate.; |
Q 3. The sum of the masses of reactants and products is equal in any physical or chemical reaction. This is in accordance with law of:; |
(A) Multiple proportion; |
(B) Definite composition; |
(C) Conservation of mass; |
(D) Reciprocal proportion; |
Q 4. If the density of a solution is \(3 12 \mathrm{~g} \mathrm{~mL}^{-1}\), the mass of \(1 5 \mathrm{~mL}\) solution in significant figures is \(\qquad\); |
(a) \(4.7 \mathrm{~g}\); |
(b) \(4680 \times 10^{-3} \mathrm{~g}\); |
(c) \(4.680 \mathrm{~g}\); |
(d) \(46.80 \mathrm{~g}\); |
Q 5. The law of definite proportions was given by -; |
(a) John Dalton; |
(b) Humphry Davy; |
(c) Proust; |
(d) Michael Faraday; |
Q 6. What is the mass of an atom of oxygen (in gm)?; |
(a) \(2.656 \times 10^{-23}\); |
(b) \(1.567 \times 10^{-22}\); |
(c) \(2.0 \times 10^{-22}\); |
(d) \(3.5 \times 10^{-23}\); |
Q 7. What is the molarity of \(0 2 \mathrm{~N} \mathrm{Na}_{2} \mathrm{CO}_{3}\) solution?; |
(a) \(0.1 \mathrm{M}\); |
(b) \(0 \mathrm{M}\); |
(c) \(0.4 \mathrm{M}\); |
(d) \(0.2 \mathrm{M}\); |
Q 8. The weight of one molecule of a compound \(\mathrm{C}_{60} \mathrm{H}_{122}\) is; |
(a) \(1.2 \times 10^{-20}\) gram; |
(b) \(1.4 \times 10^{-21}\) gram; |
(c) \(5.025 \times 10^{23} \mathrm{gram}\); |
(d) \(6.023 \times 10^{23}\) gram; |
Q 9. \(6 02 \times 10^{20}\) molecules of urea are present in \(100 \mathrm{ml}\) of its solution The concentration of urea solution is (Avogadro constant, \(\mathrm{N}_{\mathrm{A}}=6 02 \times 10^{23} \mathrm{~mol}^{-1}\) ); |
(a) \(0.02 \mathrm{M}\); |
(b) \(0.01 \mathrm{M}\); |
(c) \(0.001 \mathrm{M}\); |
(d) \(0.1 \mathrm{M}\); |
Q 10. A gaseous hydrocarbon gives upon combustion \(0 72 \mathrm{~g}\) of water and \(3 08 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) The empirical formula of the hydrocarbon is :; |
(a) \(\mathrm{C}_{2} \mathrm{H}_{4}\); |
(b) \(\mathrm{C}_{3} \mathrm{H}_{4}\); |
(c) \(\mathrm{C}_{6} \mathrm{H}_{5}\); |
(d) \(\mathrm{C}_{7} \mathrm{H}_{8}\); |
Percentage Composition
Percentage composition is a fundamental concept in chemistry that describes the relative mass of each element in a compound. It is essential for understanding the composition of compounds and is widely used in stoichiometry, chemical analysis, and various industrial applications. Here’s a detailed overview:
1. Definition
Percentage Composition: The percentage composition of a compound is the ratio of the mass of each element to the total mass of the compound, multiplied by 100.
\([ \text{Percentage Composition of Element} = \left( \frac{\text{Mass of Element in Compound}}{\text{Total Mass of Compound}} \right) \times 100 ]\)
2. Calculation of Percentage Composition
To calculate the percentage composition of an element in a compound, follow these steps:
Example: Calculate the percentage composition of water (H₂O).
3. Empirical and Molecular Formulas
Empirical Formula: The empirical formula represents the simplest whole-number ratio of atoms in a compound.
Molecular Formula: The molecular formula represents the actual number of atoms of each element in a molecule.
Determining Empirical Formula:
Example: A compound contains 40% carbon, 6.67% hydrogen, and 53.33% oxygen by mass. Determine its empirical formula.
4. Practical Applications
Chemical Analysis:
Pharmaceuticals:
Environmental Science:
5. Industrial Applications
Chemical Manufacturing:
Food Industry:
Agriculture:
Summary
Understanding percentage composition is essential for analyzing and verifying the composition of compounds. It is widely used in various fields, including chemical analysis, pharmaceuticals, environmental science, and industrial applications.
Q 1. Choose the correct statement; |
(a) The particle \(s\) in liquids are more closely held than gases but less free to move than solids.; |
(b) The particles of solids are arranged in orderly fashion but they can move as freely as liquids.; |
(c) The particles of gases are far apart as compared to solids and liquids and their movement is easy and fast.; |
(d) The particles of gases moves faster than liquids only when the gases are heated.; |
Q 2. 0 00016 can be written as A in scientific notaiton Here, A refers to; |
(a) \(1.6 \times 10^{-4}\); |
(b) \(24.50 \times 10^{-9}\); |
(c) \(2.450 \times 10^{-8}\); |
(d) \(24.50 \times 10^{-7}\); |
Q 3. A \(0.50 \mathrm{~g}\) mixture of \(\mathrm{Cu}_{2} \mathrm{O}\) and \(\mathrm{CuO}\) contains \(0.425 \mathrm{~g}\) of \(\mathrm{Cu}\). What is the mass of \(\mathrm{CuO}\) (in g) in the mixture?\n\[\n(\mathrm{Cu}=63.5 \mathrm{u}, \mathrm{O}=16 \mathrm{u})\n\]; |
(D) 0.23; |
(C) 0.27; |
(B)0.22; |
(A) 0.21; |
Q 4. In which of the following number all zeros are significant?; |
(a) 0.0005; |
(b) 0.0500; |
(c) 50.000; |
(d) 0.0050; |
Q 5. Which one of the following pairs of compounds illustrate the law of multiple proportions?; |
(a) \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{Na}_{2} \mathrm{O}\); |
(b) \(\mathrm{MgO}\) and \(\mathrm{Na}_{2} \mathrm{O}\); |
(c) \(\mathrm{Na}_{2} \mathrm{O}\) and \(\mathrm{BaO}\); |
(d) \(\mathrm{SnCl}_{2}\) and \(\mathrm{SnCl}_{4}\); |
Q 6. If the mass of the one atom is found to be \(2 324784 \times 10^{-23} \mathrm{~g}\), then this atom can be ?; |
(a) Oxygen; |
(b) Carbon; |
(c) Fluorine; |
(d) Nitrogen; |
Q 7. The molar solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is equal to :; |
(a) \(\mathrm{N} / 2\) solution; |
(b) \(\mathrm{N}\) solution; |
(c) \(2 \mathrm{~N}\) solution; |
(d) \(3 \mathrm{~N}\) solution; |
Q 8. The simplest formula of a compound containing \(50 \%\) of element \(\mathrm{X}\) (atomic mass 10 ) and \(50 \%\) of element \(\mathrm{Y}\) (atomic mass 20) is; |
(a) \(\mathrm{XY}\); |
(b) \(\mathrm{XY}_{3}\); |
(c) \(\mathrm{X}_{2} \mathrm{Y}\); |
(d) \(\mathrm{X}_{2} \mathrm{Y}_{3}\); |
Q 9. Two solutions of a substance (non electrolyte) are mixed in the following manner. \(480 \mathrm{ml}\) of \(1.5 \mathrm{M}\) first solution \(+520 \mathrm{ml}\) of \(1.2 \mathrm{M}\) second solution. What is the molarity of the final mixture?; |
(a) \(2.70 \mathrm{M}\); |
(b) \(1.344 \mathrm{M}\); |
(c) \(1.50 \mathrm{M}\); |
(d) \(1.20 \mathrm{M}\); |
Q 10. Which of the following is the correct empirical and molecular formulae of a compound, if the molecular mass of a compound is 80 and compound contains \(60 \%\) of C, \(5 \%\) of \(\mathrm{H}\) and \(35 \%\) of \(\mathrm{N}\) ?; |
(a) \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{~N} ; \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{~N}_{2}\); |
(b) \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{~N}_{2} ; \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{~N}_{4}\); |
(c) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{~N}_{2} ; \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{~N}_{4}\); |
(d) \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{~N} ; \mathrm{C}_{2} \mathrm{H}_{2} \mathrm{~N}\); |
Stoichiometry and Stoichiometric Calculations
Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It is essential for understanding how much of each substance is involved in a reaction and for predicting the amounts of products formed. Here’s a detailed overview:
1. Definition
Stoichiometry: The calculation of reactants and products in chemical reactions using the balanced chemical equation.
2. Stoichiometric Coefficients
Stoichiometric Coefficients: The numbers in front of the chemical formulas in a balanced chemical equation. They indicate the relative amounts of reactants and products.
Example: \([ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} ]\)
3. Mole Concept in Stoichiometry
Mole Concept: One mole of any substance contains \(( 6.022 \times 10^{23} )\) entities (atoms, molecules, ions, etc.), known as Avogadro’s number.
Molar Mass: The mass of one mole of a substance, expressed in grams per mole (g/mol).
Example:
4. Balancing Chemical Equations
Balancing Equations: Ensuring that the number of atoms of each element is the same on both sides of the equation.
Steps to Balance Equations:
Example: \([ \text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 ] Balanced: [ 4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 ]\)
5. Stoichiometric Calculations
Using Moles in Calculations:
Example:
Using Stoichiometric Ratios:
Example: \([ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} ]\)
6. Limiting Reactant and Excess Reactant
Limiting Reactant: The reactant that is completely consumed in a reaction, limiting the amount of product formed.
Excess Reactant: The reactant that is not completely consumed in a reaction.
Steps to Identify Limiting Reactant:
Example: \([ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 ]\)
7. Yield Calculations
Theoretical Yield: The maximum amount of product that can be formed from the given amounts of reactants.
Actual Yield: The amount of product actually obtained from a reaction.
Percent Yield: \([ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 ]\)
Example:
8. Practical Applications
Chemical Manufacturing:
Pharmaceuticals:
Environmental Science:
Summary
Stoichiometry and stoichiometric calculations are essential for understanding and predicting the outcomes of chemical reactions.
Q 1. A mixture contains two or more substances in \(\qquad\) which are called its \(\qquad\); |
(a) fixed ratio, compounds; |
(b) fixed ratio, elements; |
(c) any ratio, components; |
(d) any ratio, elements; |
Q 2. Which of the following substances cannot be separated in to its constituents by physical methods?; |
(a) Sugar and water solution; |
(b) Salt and sugar; |
(c) Solid glucose; |
(d) Both (a) and (b); |
Q 3. \(10 \mathrm{~mL}\) of gaseous hydrocarbon on combustion, gives \(40 \mathrm{~mL}\) of \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(50 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) (vapour). The hydrocarbon is:; |
(A) \(\mathrm{C}_{4} \mathrm{H}_{5}\); |
(B) \(\mathrm{C}_{8} \mathrm{H}_{10}\); |
(C) \(\mathrm{C}_{4} \mathrm{H}_{8}\); |
(D) \(\mathrm{C}_{4} \mathrm{H}_{10}\); |
Q 4. The correctly reported answer of addition of \(29 4406,3 2\) and 2 25 will have significant figures; |
(a) 3; |
(b) 4; |
(c) 2; |
(d) 5; |
Q 5. The molecular weight of \(\mathrm{O}_{2}\) and \(\mathrm{SO}_{2}\) are 32 and 64 respectively At \(15^{\circ} \mathrm{C}\) and \(150 \mathrm{~mm} \mathrm{Hg}\) pressure, one litre of \(\mathrm{O}_{2}\) contains ' \(\mathrm{N}\) ' molecules The number of molecules in two litres of \(\mathrm{SO}_{2}\) under the same conditions of temperature and pressure will be :; |
(a) \(\mathrm{N} / 2\); |
(b) \(1 \mathrm{~N}\); |
(c) \(2 \mathrm{~N}\); |
(d) \(4 \mathrm{~N}\); |
Q 6. What is the mass of 1 molecule of \(\mathrm{CO}\); |
(a) \(2.325 \times 10^{-23}\); |
(b) \(4.65 \times 10^{-23}\); |
(c) \(3.732 \times 10^{-23}\); |
(d) \(2.895 \times 10^{-23}\); |
Q 7. Volume of water needed to mix with \(10 \mathrm{~mL} 10 \mathrm{~N} \mathrm{HNO}_{3}\) to get \(0 1 \mathrm{NHNO}_{3}\) is :; |
(a) \(1000 \mathrm{~mL}\); |
(b) \(990 \mathrm{~mL}\); |
(c) \(1010 \mathrm{~mL}\); |
(d) \(10 \mathrm{~mL}\); |
Q 8. Empirical formula of hydrocarbon containing \(80 \%\) carbon and \(20 \%\) hydrogen is :; |
(a) \(\mathrm{CH}_{3}\); |
(b) \(\mathrm{CH}_{4}\); |
(c) \(\mathrm{CH}\); |
(d) \(\mathrm{CH}_{2}\); |
Q 9. Which of the following statements are correct? (i) Both solids and liquids have definite volume. (ii) Both liquids and gases do not have definite shape. (iii) Both solids and gases take the shape of the container.; |
(a) (i) and (iii); |
(b) (ii) and (iii); |
(c) (i) and (ii); |
(d) (i), (ii) and (iii); |
Q 10. Assertion : Significant figures for 0 200 is 3 whereas for 200 it is 1 Reason : Zero at the end or right of a number are significant provided they are not on the right side of the decimal point.; |
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
(c) Assertion is correct, reason is incorrect; |
(d) Assertion is incorrect, reason is correct.; |
Concentration of Solutions
Concentration of solutions is a crucial concept in chemistry that describes the amount of solute present in a given quantity of solvent or solution. Understanding this concept is essential for solving problems related to solution chemistry, stoichiometry, and various industrial applications. Here’s a detailed overview:
1. Definition
Concentration: The amount of solute present in a given quantity of solvent or solution.
2. Ways to Express Concentration
There are several ways to express the concentration of a solution, each useful for different types of calculations and applications:
a. Molarity (M)
Definition: Molarity is the number of moles of solute per liter of solution.
\([ \text{Molarity (M)} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} ]\)
Example:
b. Molality (m)
Definition: Molality is the number of moles of solute per kilogram of solvent.
\([ \text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} ]\)
Example:
c. Mole Fraction (χ)
Definition: Mole fraction is the ratio of the number of moles of a component to the total number of moles of all components in the solution.
\([ \text{Mole Fraction (χ)} = \frac{\text{Number of moles of component}}{\text{Total number of moles of all components}} ]\)
Example:
d. Mass Percent (Weight Percent)
Definition: Mass percent is the mass of solute divided by the total mass of the solution, multiplied by 100.
\([ \text{Mass Percent} = \left( \frac{\text{Mass of solute}}{\text{Total mass of solution}} \right) \times 100 ]\)
Example:
e. Volume Percent
Definition: Volume percent is the volume of solute divided by the total volume of the solution, multiplied by 100.
\([ \text{Volume Percent} = \left( \frac{\text{Volume of solute}}{\text{Total volume of solution}} \right) \times 100 ]\)
Example:
3. Dilution of Solutions
Dilution: The process of reducing the concentration of a solute in a solution, usually by adding more solvent.
Dilution Formula: \([ M_1 V_1 = M_2 V_2 ]\) where \(( M_1 )\) and \(( M_2 )\) are the initial and final molarities, and \(( V_1 )\) and \(( V_2 )\) are the initial and final volumes.
Example:
4. Practical Applications
Chemical Reactions:
Pharmaceuticals:
Environmental Science:
5. Industrial Applications
Chemical Manufacturing:
Food Industry:
Agriculture:
Summary
Understanding the concentration of solutions is essential for analyzing and controlling the composition of solutions in various fields. It is widely used in chemical reactions, pharmaceuticals, environmental science, and industrial applications.
Q 1. Which one of these is not a pure compound?; |
(a) \(\mathrm{O}_{3}\); |
(b) \(\mathrm{H}_{2} \mathrm{O}_{2}\); |
(c) \(\mathrm{H}_{2} \mathrm{O}\); |
(d) Sucrose solution; |
Q 2. Which of the following is an example of a heterogeneous substance?; |
(a) Bottled water; |
(b) Table salt; |
(c) Pieces of copper; |
(d) Candle; |
Q 3. The empirical formula of a hydrocarbon containing \(80 \%\) carbon and \(20 \%\) hydrogen is:; |
(A) \(\mathrm{CH}\); |
(B) \(\mathrm{CH}_{2}\); |
(C) \(\mathrm{CH}_{3}\); |
(D) \(\mathrm{CH}_{4}\); |
Q 4. The number of significant figures in \(10 3106 \mathrm{~g}\) is; |
(a) 2; |
(b) 3; |
(c) 1; |
(d) 6; |
Q 5. Number of \(g\) of oxygen in \(32 2 \mathrm{~g} \mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}\) is; |
(a) 20.8; |
(b) 2.24; |
(c) 22.4; |
(d) 2.08; |
Q 6. Calculate the volume at STP occupied by \(240 \mathrm{gm}\) of \(\mathrm{SO}_{2}\); |
(a) 64; |
(b) 84; |
(c) 59; |
(d) 73; |
Q 7. One kilogram of a sea water sample contains \(6 \mathrm{mg}\) of dissolved \(\mathrm{O}_{2}\) The concentration of \(\mathrm{O}_{2}\) in the sample in ppm is; |
(a) 0.6; |
(b) 6.0; |
(c) 60.0; |
(d) 16.0; |
Q 8. The empirical formula of a compound is \(\mathrm{CH}_{2}\) One mole of this compound has a mass of 42 grams Its molecular formula is :; |
(a) \(\mathrm{C}_{3} \mathrm{H}_{6}\); |
(b) \(\mathrm{C}_{3} \mathrm{H}_{8}\); |
(c) \(\mathrm{CH}_{2}\); |
(d) \(\mathrm{C}_{2} \mathrm{H}_{2}\); |
Q 9. Choose correct option based on following statements Here \(\mathrm{T}\) stands for true statement and \(\mathrm{F}\) for false statement (i) Homogeneous mixture has uniform composition throughout. (ii) All components of a heterogeneous mixture are observable to naked eyes. (iii) All solutions are homogeneous in nature. (iv) Air is an example of heterogeneous mixture.; |
(a) TTFF; |
(b) TFTF; |
(c) FFTT; |
(d) TFFF; |
Q 10. Assertion : 1 231 has three significant figures Reason : All numbers right to the decimal point are significant.; |
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
(c) Assertion is correct, reason is incorrect; |
(d) Assertion is incorrect, reason is correct.; |
Limiting Reagent
The concept of the limiting reagent is crucial in stoichiometry and chemical reactions. It helps determine the amount of product formed in a reaction and identifies which reactant will be completely consumed first. Here’s a detailed overview:
1. Definition
Limiting Reagent: The reactant that is completely consumed in a chemical reaction, limiting the amount of product formed. It determines the maximum amount of product that can be produced.
Excess Reagent: The reactant that is not completely consumed in a reaction and remains after the reaction is complete.
2. Identifying the Limiting Reagent
To identify the limiting reagent in a chemical reaction, follow these steps:
Example: Consider the reaction between hydrogen and oxygen to form water:\( [ 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} ]\)
Suppose we have 10 moles of \((\text{H}_2) and 7 moles of (\text{O}_2)\).
3. Calculating the Amount of Product
Once the limiting reagent is identified, use it to calculate the amount of product formed.
Example: Using the previous example, calculate the amount of water \(((\text{H}_2\text{O}))\) formed.
4. Practical Applications
Chemical Manufacturing:
Pharmaceuticals:
Environmental Science:
5. Industrial Applications
Chemical Manufacturing:
Food Industry:
Agriculture:
Summary
Understanding the concept of the limiting reagent is essential for predicting the outcomes of chemical reactions and optimizing the use of reactants. It is widely used in chemical manufacturing, pharmaceuticals, environmental science, and various industrial applications.
Q 1. One fermi is; |
(a) \(10^{-15} \mathrm{~cm}\); |
(b) \(10^{-13} \mathrm{~cm}\); |
(c) \(10^{-10} \mathrm{~cm}\); |
(d) \(10^{-12} \mathrm{~cm}\); |
Q 2. Difference in density is the basis of; |
(a) ultrafiltration; |
(b) molecular sieving; |
(c) molecular attraction; |
(d) gravity separation; |
Q 3. What percentage of oxygen is present in the compound \(\mathrm{CaCO}_{3} \cdot 3 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) ?; |
(A) \(40.54 \%\); |
(B) \(35.56 \%\); |
(C) \(15.40 \%\); |
(D) \(41.95 \%\); |
Q 4. Choose the correct option that represents the result of the given calculation to the appropriate number of significant figures: \[ \frac{43.0 \times 0.0243}{0.340 \times 4} \]; |
(a) 0.768; |
(b) 0.77; |
(c) 0.76; |
(d) 0.7683; |
Q 5. The number of moles of oxygen in one litre of air containing \(21 \%\) oxygen by volume, under standard conditions are; |
(a) 0.0093 mole; |
(b) 0.21 mole; |
(c) 2.10 mole; |
(d) 0.186 mole; |
Q 6. At S T P the density of \(\mathrm{CCl}_{4}\) vapours in \(\mathrm{g} / \mathrm{L}\) will be nearest to:; |
(a) 6.87; |
(b) 3.42; |
(c) 10.26; |
(d) 4.57; |
Q 7. The mass of a molecule of water is; |
(a) \(3 \times 10^{-25} \mathrm{~kg}\); |
(b) \(3 \times 10^{-26} \mathrm{~kg}\); |
(c) \(1.5 \times 10^{-26} \mathrm{~kg}\); |
(d) \(2.5 \times 10^{-26} \mathrm{~kg}\); |
Q 8. A compound contains \(54 55 \%\) carbon, \(9 09 \%\) hydrogen , \(36 36 \%\) oxygen The empirical formula of this compound is; |
(a) \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}\); |
(b) \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\); |
(c) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\); |
(d) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\); |
Q 9. Read the following and choose the incorrect statements (i) Both weight and mass are same quantities used for measurement of amount of matter present in a substance (ii) Mass and weight of a substance vary from one place to another due to change in gravity. (iii) SI unit of mass is kilogram and while SI unit of weight is gram.; |
(a) (i) and (iii); |
(b) (ii) and (iii); |
(c) (i) and (ii); |
(d) All of these; |
Q 10. Assertion : One atomic mass unit is defined as one twelfth of the mass of one carbon - 12 atom Reason : Carbon-12 isotope is the most abundunt isotope of carbon and has been chosen as standard.; |
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
(c) Assertion is correct, reason is incorrect; |
(d) Assertion is incorrect, reason is correct.; |
Empirical and Molecular Formulas
Understanding empirical and molecular formulas is essential for determining the composition of compounds and solving problems in stoichiometry. Here’s a detailed overview:
1. Definitions
Empirical Formula: The empirical formula of a compound gives the simplest whole-number ratio of the atoms of each element present in the compound.
Molecular Formula: The molecular formula of a compound gives the actual number of atoms of each element present in a molecule of the compound. It is a multiple of the empirical formula.
2. Determining Empirical Formula
To determine the empirical formula of a compound, follow these steps:
Example: A compound contains 40% carbon, 6.67% hydrogen, and 53.33% oxygen by mass. Determine its empirical formula.
3. Determining Molecular Formula
To determine the molecular formula, follow these steps:
Example: The empirical formula of a compound is \(( \text{CH}_2\text{O} )\) and its molar mass is 180 g/mol. Determine the molecular formula.
4. Practical Applications
Chemical Analysis:
Pharmaceuticals:
Environmental Science:
5. Industrial Applications
Chemical Manufacturing:
Food Industry:
Agriculture:
Summary
Understanding empirical and molecular formulas is essential for analyzing and verifying the composition of compounds. It is widely used in chemical analysis, pharmaceuticals, environmental science, and industrial applications.
Q 1. The prefix \(10^{18}\) is; |
(a) giga; |
(b) kilo; |
(c) exa; |
(d) nano; |
Q 2. A mixture of sand and iodine can be separated by; |
(a) crystallisation; |
(b) distillation; |
(c) sublimation; |
(d) fractionation; |
Q 3. Find out the normality of a solution, when \(9.8 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is dissolved in \(500 \mathrm{~mL}\) of solution.; |
(A) 0.4; |
(B) 0.2; |
(C) 0.8; |
(D) 4.0; |
Q 4. Arrange the numbers in increasing no of significant figures \(0 002600,2 6000,2 6,0 260\); |
(a) \(2.6<0.260<0.002600<2.6000\); |
(b) \(2.6000<2.6<0.002600<0.260\); |
(c) \(0.260<2.6<0.002600<2.6000\); |
(d) \(0.002600<0.260<2.6<2.6000\); |
Q 5. \(10 \mathrm{dm}^{3}\) of \(\mathrm{N}_{2}\) gas and \(10 \mathrm{dm}^{3}\) of gas \(\mathrm{X}\) at the same temperature contain the same number of molecules, the gas \(\mathrm{X}\) is; |
(a) \(\mathrm{CO}_{2}\); |
(b) \(\mathrm{CO}\); |
(c) \(\mathrm{H}_{2}\); |
(d) \(\mathrm{NO}\); |
Q 6. The number of gram molecules of oxygen in \(6 02 \times 10^{24}\) CO molecules is; |
(a) 10 gm molecules; |
(b) \(5 \mathrm{gm}\) molecules; |
(c) 1 gm molecules; |
(d) 0.5 gm molelcules; |
Q 7. One mole of \(\mathrm{CO}_{2}\) contains :; |
(a) \(3 \mathrm{~g}\) atoms of \(\mathrm{CO}_{2}\); |
(b) \(18.1 \times 10^{23}\) molecules of \(\mathrm{CO}_{2}\); |
(c) \(6.02 \times 10^{23}\) atoms of \(\mathrm{O}\); |
(d) \(6.02 \times 10^{23}\) atoms of \(\mathrm{C}\); |
Q 8. In a hydrocarbon, mass ratio of hydrogen and carbon is \(1: 3\), the empirical formula of hydrocarbon is; |
(a) \(\mathrm{CH}_{4}\); |
(b) \(\mathrm{CH}_{2}\); |
(c) \(\mathrm{C}_{2} \mathrm{H}\); |
(d) \(\mathrm{CH}_{3}\); |
Q 9. Moon takes 27 3 days to complete one orbit around the Earth Now read the following statements and choose the correct code Here \(\mathrm{T}\) is for true statement and \(\mathrm{F}\) is for 'False statement' (i) Moon takes 655.2 hours to complete one orbit around the Earth. (ii) Moon takes 39312 seconds to complete one orbit around the earth. (iii) Moon takes 1638 minutes to complete one orbit around the Earth.; |
(a) F T F; |
(b) \(\mathrm{T} \mathrm{T} \mathrm{T}\); |
(c) T F F; |
(d) TF T; |
Q 10. Assertion : Volume of a gas is inversely proportional to the number of moles of gas Reason : The ratio by volume of gaseous reactants and products is in agreement with their mole ratio.; |
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
(c) Assertion is correct, reason is incorrect; |
(d) Assertion is incorrect, reason is correct.; |
Avogadro’s Law
Definition: Avogadro’s Law states that the volume of a gas is directly proportional to the number of moles of the gas, provided the temperature and pressure remain constant. Mathematically, it is expressed as: \([ V \propto n ]\) or \([ \frac{V}{n} = k ]\) where ( V ) is the volume of the gas, ( n ) is the number of moles, and ( k ) is a constant.
Formula: \([ \frac{V_1}{n_1} = \frac{V_2}{n_2} ]\) This equation shows that if the amount of gas changes, the volume changes proportionally, assuming constant temperature and pressure.
Graphical Representation: A graph of volume (( V )) versus the number of moles (( n )) at constant temperature and pressure is a straight line passing through the origin, indicating direct proportionality.
Derivation from Ideal Gas Law
The ideal gas law is given by: [ PV = nRT ] where ( P ) is the pressure, ( V ) is the volume, ( n ) is the number of moles, ( R ) is the universal gas constant, and ( T ) is the temperature.
At constant ( P ) and ( T ): \([ V = \frac{nRT}{P} ]\) Since \(( \frac{RT}{P} )\) is a constant, we can write: \([ V \propto n ]\)
Molar Volume
At standard temperature and pressure (STP: 0°C and 1 atm), one mole of any gas occupies 22.4 liters. This is known as the molar volume of a gas.
Real-Life Examples
Industrial Applications
Practical Applications
Summary
Avogadro’s Law is fundamental in understanding the behavior of gases. It is crucial for various scientific and industrial applications, from breathing and balloon inflation to large-scale chemical manufacturing and environmental monitoring.
Q 1. The prefix zepto stands for (in \(\mathrm{m}\) ); |
(a) \(10^{9}\); |
(b) \(10^{-12}\); |
(c) \(10^{-15}\); |
(d) \(10^{-21}\); |
Q 2. A \(0.50 \mathrm{~g}\) mixture of \(\mathrm{Cu}_{2} \mathrm{O}\) and \(\mathrm{CuO}\) contains \(0.425 \mathrm{~g}\) of \(\mathrm{Cu}\). What is the mass of \(\mathrm{CuO}\) (in g) in the mixture?\n\[\n(\mathrm{Cu}=63.5 \mathrm{u}, \mathrm{O}=16 \mathrm{u})\n\]; |
(A) 0.21; |
(B)0.22; |
(C) 0.27; |
(D) 0.23; |
Q 3. Mixture of \(1 \mathrm{~g}\) each of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{NaHCO}_{3}\) is reacted with \(0.1 \mathrm{NHCl}\). The quantity of \(0.1 \mathrm{MHCl}\) required to react completely with the above mixture is; |
(A) \(15.78 \mathrm{~mL}\); |
(B) \(157.8 \mathrm{~mL}\); |
(C) \(198.4 \mathrm{~mL}\); |
(D) \(308 \mathrm{~mL}\); |
Q 4. Dimension of pressure are same as that of; |
(a) Energy; |
(b) Force; |
(c) Force per unit volume; |
(d) Energy per unit volume; |
Q 5. One mole of a gas occupies a volume of \(22 4 \mathrm{~L}\) This is derived from; |
(a) Berzelius' hypothesis; |
(b) Gay-Lussac's law; |
(c) Avogadro's law; |
(d) Dalton's law; |
Q 6. The number of oxygen atoms in \(4 4 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) is; |
(a) \(1.2 \times 10^{23}\); |
(b) \(6 \times 10^{22}\); |
(c) \(6 \times 10^{23}\); |
(d) \(12 \times 10^{23}\); |
Q 7. Volume of a gas at NTP is \(1 12 \times 10^{-7} \mathrm{~cm}^{3}\) The number of molecules in it is :; |
(a) \(3.01 \times 10^{12}\); |
(b) \(3.01 \times 10^{24}\); |
(c) \(3.01 \times 10^{23}\); |
(d) \(3.01 \times 10^{20}\); |
Q 8. An organic compound contains carbon, hydrogen and oxygen Its elemental analysis gave C, \(38 71 \%\) and \(\mathrm{H}, 9 67 \%\) The empirical formula of the compound would be :; |
(a) \(\mathrm{CH}_{3} \mathrm{O}\); |
(b) \(\mathrm{CH}_{2} \mathrm{O}\); |
(c) \(\mathrm{CHO}\); |
(d) \(\mathrm{CH}_{4} \mathrm{O}\); |
Q 9. Give the correct order of initials \(\mathrm{T}\) or \(\mathrm{F}\) for following statements Use T if statement is true and F if it is false (i) Gay-Lussac's law of gaseous volumes is actually the law of definite proportion by volume. (ii) Law of conservation of mass is true for physical change, but not for chemical change. (iii) The percentage of oxygen in \(\mathrm{H}_{2} \mathrm{O}_{2}\) is different from that in \(\mathrm{H}_{2} \mathrm{O}\). Hence, it violates law of definite proportions. (iv) Fixed mass of A reacts with two different masses of \(\mathrm{B}\) (say \(x\) and \(y\) ), then the ratio of \(x / y\) can be any positive integer. (v) At STP, \(5 \mathrm{~mL}\) of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) have different no. of molecules.; |
(a) TTFTF; |
(b) FTTFT; |
(c) TFFTF; |
(d) TFTTF; |
Q 10. Assertion : Equal moles of different substances contain same number of constituent particles Reason : Equal weights of different substances contain the same number of constituent particles.; |
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
(c) Assertion is correct, reason is incorrect; |
(d) Assertion is incorrect, reason is correct.; |
Gay-Lussac’s Law
Definition: Gay-Lussac’s Law states that the pressure of a given mass of gas varies directly with the absolute temperature when the volume is kept constant. Mathematically, it is expressed as: \([ P \propto T ]\) or \([ \frac{P}{T} = k ]\) where ( P ) is the pressure, ( T ) is the absolute temperature, and ( k ) is a constant.
Formula: \([ \frac{P_1}{T_1} = \frac{P_2}{T_2} ]\) This equation shows that if the temperature of a gas changes, the pressure changes proportionally, assuming constant volume.
Graphical Representation: A graph of pressure (( P )) versus temperature (( T )) at constant volume is a straight line passing through the origin, indicating direct proportionality.
Real-Life Examples:
Industrial Applications:
Q 1. The unit \(\mathrm{J} \mathrm{Pa}^{-1}\) is equivalent to; |
(a) \(\mathrm{m}^{3}\); |
(b) \(\mathrm{cm}^{3}\); |
(c) \(\mathrm{dm}^{3}\); |
(d) None of these; |
Q 2. \(10 \mathrm{~mL}\) of gaseous hydrocarbon on combustion, gives \(40 \mathrm{~mL}\) of \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(50 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) (vapour). The hydrocarbon is:; |
(A) \(\mathrm{C}_{4} \mathrm{H}_{5}\); |
(B) \(\mathrm{C}_{8} \mathrm{H}_{10}\); |
(C) \(\mathrm{C}_{4} \mathrm{H}_{8}\); |
(D) \(\mathrm{C}_{4} \mathrm{H}_{10}\); |
Q 3. When \(25 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is dissolved in \(10^{3} \mathrm{Kg}\) of solution, its concentration will be; |
(A) \(2.5 \mathrm{ppm}\); |
(B) \(25 \mathrm{ppm}\); |
(C) \(250 \mathrm{ppm}\); |
(D) \(100 \mathrm{ppm}\); |
Q 4. \(n g\) of substance \(X\) reacts with \(m g\) of substance \(Y\) to form \(\mathrm{p} g\) of substance \(\mathrm{R}\) and \(\mathrm{q} g\) of substance \(\mathrm{S}\) This reaction can be represented as, \(\mathrm{X}+\mathrm{Y}=\mathrm{R}+\mathrm{S}\) The relation which can be established in the amounts of the reactants and the products will be; |
(a) \(\mathrm{n}-\mathrm{m}=\mathrm{p}-\mathrm{q}\); |
(b) \(\mathrm{n}+\mathrm{m}=\mathrm{p}+\mathrm{q}\); |
(c) \(\mathrm{n}=\mathrm{m}\); |
(d) \(\mathrm{p}=\mathrm{q}\); |
Q 5. One of the following combination which illustrates the law of reciprocal proportions?; |
(a) \(\mathrm{N}_{2} \mathrm{O}_{3}, \mathrm{~N}_{2} \mathrm{O}_{4}, \mathrm{~N}_{2} \mathrm{O}_{5}\); |
(b) \(\mathrm{NaCl}, \mathrm{NaBr}, \mathrm{NaI}\); |
(c) \(\mathrm{CS}_{2}, \mathrm{CO}_{2}, \mathrm{SO}_{2}\); |
(d) \(\mathrm{PH}_{3}, \mathrm{P}_{2} \mathrm{O}_{3}, \mathrm{P}_{2} \mathrm{O}_{5}\); |
Q 6. Which has maximum number of molecules?; |
(a) \(7 \mathrm{gm} \mathrm{N}_{2}\); |
(b) \(2 \mathrm{gm} \mathrm{H}_{2}\); |
(c) \(16 \mathrm{gm} \mathrm{NO}_{2}\); |
(d) \(16 \mathrm{gmO}_{2}\); |
Q 7. How many atoms are contained in one mole of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) ?; |
(a) \(20 \times 6.02 \times 10^{23}\) atoms \(/ \mathrm{mol}\); |
(b) \(45 \times 6.02 \times 10^{23}\) atoms \(/ \mathrm{mol}\); |
(c) \(5 \times 6.02 \times 10^{23}\) atoms \(/ \mathrm{mol}\); |
(d) None of these; |
Q 8. A hydrocarbon is composed of \(75 \%\) carbon The empirical formula of the compound is; |
(a) \(\mathrm{CH}_{2}\); |
(b) \(\mathrm{CH}_{3}\); |
(c) \(\mathrm{C}_{2} \mathrm{H}_{5}\); |
(d) \(\mathrm{CH}_{4}\); |
Q 9. Consider the following statements (i) Atoms of \(\mathrm{H}, \mathrm{O}, \mathrm{N}\) and \(\mathrm{C}\) have identical properties but different mass. (ii) Matter is divisible into atoms which are further indivisible. (iii) The ratio of \(\mathrm{N}\) : \(\mathrm{H}\) in \(\mathrm{NH}_{3}\) is \(1: 3\) and \(\mathrm{N}\) : \(\mathrm{O}\) in nitric oxide is \(2: 1\). (iv) Dalton's atomic theory support law of conservation of mass. Which of the following pairs of statements is true according to Dalton's atomic theory?; |
(a) (i) and (ii); |
(b) (ii) and (iii); |
(c) (ii) and (iv); |
(d) (i) and (iv); |
Q 10. Assertion : The empirical mass of ethene is half of its molecular mass Reason : The empirical formula represents the simplest whole number ratio of various atoms present in a compound.; |
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
(c) Assertion is correct, reason is incorrect; |
(d) Assertion is incorrect, reason is correct.; |
Law of Definite Proportions
Definition: The Law of Definite Proportions, also known as Proust’s Law, states that a chemical compound always contains exactly the same proportion of elements by mass.
Example: Water \(( \text{H}_2\text{O} )\) always contains hydrogen and oxygen in a mass ratio of 1:8, regardless of the source or method of preparation.
Historical Context: Formulated by Joseph Proust in 1799, this law was fundamental in establishing the concept of chemical compounds having fixed compositions.
Applications:
Q 1. Which has highest weight?; |
(a) \(1 \mathrm{~m}^{3}\) of water; |
(b) A normal adult man; |
(c) 10 litre of \(\mathrm{Hg}\); |
(d) All have same weight; |
Q 2. The empirical formula of a hydrocarbon containing \(80 \%\) carbon and \(20 \%\) hydrogen is:; |
(A) \(\mathrm{CH}\); |
(B) \(\mathrm{CH}_{2}\); |
(C) \(\mathrm{CH}_{3}\); |
(D) \(\mathrm{CH}_{4}\); |
Q 3. 1 litre of a hydrocarbon weighs as much as one litre of \(\mathrm{CO}_{2}\) at same \(\mathrm{P}\) and \(\mathrm{T}\). Then the molecular formula of the hydrocarbon is:; |
(A) \(\mathrm{C}_{3} \mathrm{H}_{8}\); |
(B) \(\mathrm{C}_{2} \mathrm{H}_{6}\); |
(C) \(\mathrm{C}_{2} \mathrm{H}_{4}\); |
(D) \(\mathrm{C}_{3} \mathrm{H}_{6}\); |
Q 4. \(20 \mathrm{~g}\) of \(\mathrm{CaCO}_{3}\) on heating gave \(8 8 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(11 2 \mathrm{~g}\) of \(\mathrm{CaO}\) This is in accordance with; |
(a) The law of conservation of mass.; |
(b) The law of constant composition.; |
(c) The law of reciprocal proportion.; |
(d) None of these; |
Q 5. Equal volumes of two gases \(\mathrm{A}\) and \(\mathrm{B}\) are kept in a container at the same temperature and pressure Avogadro's law is invalid if; |
(a) the gases are reactive; |
(b) the gases are non-reactive; |
(c) gas A has more number of molecules than gas B.; |
(d) None of these; |
Q 6. Number of atoms in 558 5 gram \(\mathrm{Fe}\) (at wt of \(\mathrm{Fe}=55 85\) \(\mathrm{g} \mathrm{\textrm {mol } ^ { - 1 }}\) ) is; |
(a) twice that in \(60 \mathrm{~g}\) carbon; |
(b) \(6.023 \times 10^{22}\); |
(c) half that in \(8 \mathrm{~g} \mathrm{He}\); |
(d) \(558.5 \times 6.023 \times 10^{23}\); |
Q 7. One litre oxygen gas at S T P will weigh :; |
(a) \(1.43 \mathrm{~g}\); |
(b) \(2.24 \mathrm{~g}\); |
(c) \(11.2 \mathrm{~g}\); |
(d) \(22.4 \mathrm{~g}\); |
Q 8. \(12 \mathrm{gm}\) of \(\mathrm{Mg}\) (atomic mass 24 ) will react completely with hydrochloric acid to give; |
(a) One mol of \(\mathrm{H}_{2}\); |
(b) \(1 / 2 \mathrm{~mol}\) of \(\mathrm{H}_{2}\); |
(c) \(2 / 3 \mathrm{~mol}\) of \(\mathrm{O}_{2}\); |
(d) both \(1 / 2 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) and \(1 / 2 \mathrm{~mol}\) of \(\mathrm{O}_{2}\); |
Q 9. Choose the correct option based on following statements Here ' \(T\) ' stands for true and ' \(F\) ' stands for false statement (i) Molecular mass of cane sugar \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is \(182 \mathrm{amu}\). (ii) 1 mole of cane sugar contains \(6.022 \times 10^{23}\) molecules of cane sugar. (iii) \(34.20 \mathrm{~g}\) of cane sugar contains \(6.022 \times 10^{21}\) molecules of cane sugar.; |
(a) TTF; |
(b) \(\mathrm{TFT}\); |
(c) FTF; |
(d) FTT; |
Q 10. What are the significant figure(s) in a broken "ruler" show below? (A) 1 (B) 2 (C) 3 (D) 0; |
(a) A, B and C; |
(b) A, B, D; |
(c) A only; |
(d) A and B; |
Balancing Chemical Equations
Importance: Balancing chemical equations is essential for understanding the stoichiometry of reactions, ensuring the conservation of mass and atoms.
Steps to Balance Equations:
Example: Balancing the combustion of methane: \([ \text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} ]\) Balanced Equation: \([ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} ]\)
Applications:
Q 1. Which one of the following set of units represents the smallest and largest amount of energy respectively?; |
(a) J and erg; |
(b) erg and cal; |
(c) cal and \(\mathrm{eV}\); |
(d) \(\mathrm{eV}\) and \(\mathrm{L}\)-atm; |
Q 2. What percentage of oxygen is present in the compound \(\mathrm{CaCO}_{3} \cdot 3 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) ?; |
(A) \(40.54 \%\); |
(B) \(35.56 \%\); |
(C) \(15.40 \%\); |
(D) \(41.95 \%\); |
Q 3. A complex of iron contains \(45.6 \%\) iron by mass. Find out the number of iron atoms in \(5.00 \mathrm{~g}\) of this complex number of iron atoms \(=x^{*} 10^{\wedge} 22\) find the \(x\); |
(A) 2.1; |
(B) 2.2; |
(C) 2.3; |
(D) 2.4; |
Q 4. Which of the following is the best example of law of conservation of mass?; |
(a) \(12 \mathrm{~g}\) of carbon combines with \(32 \mathrm{~g}\) of oxygen to form \(44 \mathrm{~g}\) of \(\mathrm{CO}_{2}\); |
(b) When \(12 \mathrm{~g}\) of carbon is heated in a vacuum there is no change in mass; |
(c) A sample of air increases in volume when heated at constant pressure but its mass remains unaltered; |
(d) The weight of a piece of platinum is the same before and after heating in air; |
Q 5. Molecular mass is defined as the; |
(a) mass of one atom compared with the mass of one molecule; |
(b) mass of one atom compared with the mass of one atom of hydrogen; |
(c) mass of one molecule of any substance compared with the mass of one atom of C-12; |
(d) None of the above; |
Q 6. The number of molecules in \(16 \mathrm{~g}\) of methane is; |
(a) \(3.0 \times 10^{23}\); |
(b) \(\frac{16}{6.02} \times 10^{23}\); |
(c) \(6.023 \times 10^{23}\); |
(d) \(\frac{16}{3.0} \times 10^{23}\); |
Q 7. Number of moles of \(\mathrm{NaOH}\) present in 2 litre of \(0 5 \mathrm{M} \mathrm{NaOH}\) is :; |
(a) 1.5; |
(b) 2.0; |
(c) 1.0; |
(d) 2.5; |
Q 8. \(20 0 \mathrm{~kg}\) of \(\mathrm{N}_{2(\mathrm{~g})}\) and \(3 0 \mathrm{~kg}\) of \(\mathrm{H}_{2(\mathrm{~g})}\) are mixed to produce \(\mathrm{NH}_{3(\mathrm{~g})}\) The amount of \(\mathrm{NH}_{3(\mathrm{~g})}\) formed is; |
(a) \(17 \mathrm{~kg}\); |
(b) \(34 \mathrm{~kg}\); |
(c) \(20 \mathrm{~kg}\); |
(d) \(3 \mathrm{~kg}\); |
Q 9. How much time (in hours) would it take to distribute one Avogadro number of wheat grains if \(10^{20}\) grains are distributed each second ?; |
(a) 0.1673; |
(b) 1.673; |
(c) 16.73; |
(d) 167.3; |
Q 10. Which one of the following sets of compounds correctly illustrate the law of reciprocal proportions?; |
(a) \(\mathrm{P}_{2} \mathrm{O}_{3}, \mathrm{PH}_{3}, \mathrm{H}_{2} \mathrm{O}\); |
(b) \(\mathrm{P}_{2} \mathrm{O}_{5}, \mathrm{PH}_{3}, \mathrm{H}_{2} \mathrm{O}\); |
(c) \(\mathrm{N}_{2} \mathrm{O}_{5}, \mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{O}\); |
(d) \(\mathrm{N}_{2} \mathrm{O}, \mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{O}\); |
States of Matter
Definition: Matter exists in three primary states: solid, liquid, and gas. Each state has distinct physical properties.
Properties:
Phase Transitions:
Applications:
Q 1. A measured temperature on Fahrenheit scale is \(200^{\circ} \mathrm{F}\) What will this reading be on Celsius scale?; |
(a) \(40^{\circ} \mathrm{C}\); |
(b) \(94^{\circ} \mathrm{C}\); |
(c) \(93.3^{\circ} \mathrm{C}\); |
(d) \(30^{\circ} \mathrm{C}\); |
Q 2. Find out the normality of a solution, when \(9.8 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is dissolved in \(500 \mathrm{~mL}\) of solution.; |
(A) 0.4; |
(B) 0.2; |
(C) 0.8; |
(D) 4.0; |
Q 3. The sum of the masses of reactants and products is equal in any physical or chemical reaction. This is in accordance with law of:; |
(A) Multiple proportion; |
(B) Definite composition; |
(C) Conservation of mass; |
(D) Reciprocal proportion; |
Q 4. Which of the following statements is correct about the reaction given below? \(4 \mathrm{Fe}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~g})\); |
(a) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore, it follows law of conservation of mass.; |
(b) Total mass of reactants \(=\) total mass of product; therefore, law of multiple proportions is followed.; |
(c) Amount of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) can be increased by reducing the amount of any one of the reactants (iron or oxygen).; |
(d) Amount of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.; |
Q 5. \(1 \mathrm{amu}\) is equal to; |
(a) \(\frac{1}{14}\) ofO-16; |
(b) \(\frac{1}{12}\) of C-12; |
(c) \(1 \mathrm{~g}\) of \(\mathrm{H}_{2}\); |
(d) \(1.66 \times 10^{-23} \mathrm{~kg}\); |
Q 6. Among the following pairs of compounds, the one that illustrates the law of multiple proportions is; |
(a) \(\mathrm{NH}_{3}\) and \(\mathrm{NCl}_{3}\); |
(b) \(\mathrm{H}_{2} \mathrm{~S}\) and \(\mathrm{SO}_{2}\); |
(c) \(\mathrm{CS}_{2}\) and \(\mathrm{FeSO}_{4}\); |
(d) \(\mathrm{CuO}\) and \(\mathrm{Cu}_{2} \mathrm{O}\); |
Q 7. \(\mathrm{O}_{2}, \mathrm{~N}_{2}\) are present in the ratio of \(1: 4\) by weight The ratio of number of molecules is :; |
(a) \(7: 32\); |
(b) \(1: 4\); |
(c) \(2: 1\); |
(d) \(4: 1\); |
Q 8. 20.0 kg of \(H_{2}(g)\) and 32 kg of \(O_{2}(g)\) are reacted to produce \(H_{2}O(l)\). The amount of \(\mathrm{H}_{2} \mathrm{O}\) (1) formed after completion of reaction is; |
(a) \(62 \mathrm{~kg}\); |
(b) \(38 \mathrm{~kg}\); |
(c) \(42 \mathrm{~kg}\); |
(d) \(72 \mathrm{~kg}\); |
Q 9. Arrange the following in the order of increasing mass (atomic mass: \(\mathrm{O}=16, \mathrm{Cu}=63, \mathrm{~N}=14\) ) I. one atom of oxygen II. one atom of nitrogen III. \(1 \times 10^{-10}\) mole of oxygen IV. \(1 \times 10^{-10}\) mole of copper; |
(a) II \(<\) I \(<\) III \(<\) IV; |
(b) I \(<\) II \(<\) III \(<\) IV; |
(c) III \(<\) II \(<\) IV \(<\) I; |
(d) IV \(<\) II \(<\) III \(<\) I; |
Q 10. If we consider that \(1 / 6\), in place of \(1 / 12\), mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will; |
(a) decrease twice; |
(b) increase two fold; |
(c) remain unchanged; |
(d) be a function of the molecular mass of the substance; |
Nature of Matter
Definition: Matter is anything that has mass and occupies space. It is composed of atoms and molecules.
Dual Nature: Matter exhibits both particle and wave properties, as demonstrated by the wave-particle duality in quantum mechanics.
Key Concepts:
Applications:
Quantum Computing: Utilizing quantum states of matter for computation.
Q 1. Which of the following is not a SI unit?; |
(a) metre; |
(b) candela; |
(c) mole; |
(d) litre; |
Q 2. Mixture of \(1 \mathrm{~g}\) each of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{NaHCO}_{3}\) is reacted with \(0.1 \mathrm{NHCl}\). The quantity of \(0.1 \mathrm{MHCl}\) required to react completely with the above mixture is; |
(A) \(15.78 \mathrm{~mL}\); |
(B) \(157.8 \mathrm{~mL}\); |
(C) \(198.4 \mathrm{~mL}\); |
(D) \(308 \mathrm{~mL}\); |
Q 3. If the true value for an experimental result is 6 23 and the results reported by three students \(\mathrm{X}, \mathrm{Y}\) and \(\mathrm{Z}\) are : \(\mathrm{X}: 6.18\) and 6.28 \(\mathrm{Y}: \quad 6.20\) and 6.023 \(\mathrm{Z}: \quad 6.22\) and 6.24 Which of the following option is correct :; |
(a) \(\mathrm{X}\) precise, \(\mathrm{Y}\) accurate, \(\mathrm{Z}\) precise and accurate.; |
(b) \(\mathrm{X}\) precise and accurate, \(\mathrm{Y}\) not precise, \(\mathrm{Z}\) precise; |
(c) Both \(\mathrm{X} \& \mathrm{Z}\) precise \& accurate, \(\mathrm{Y}\) not precise.; |
(d) Both \(\mathrm{X} \& \mathrm{Y}\) neither precise nor accurate, \(\mathrm{Z}\) both precise and accurate.; |
Q 4. In an experiment \(4 2 \mathrm{~g}\) of \(\mathrm{NaHCO}_{3}\) is added to a solution of acetic acid weighing \(10 0 \mathrm{~g}\), it is observed that \(2 2 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) is released into the atmosphere The residue left behind is found to weigh \(12 0 \mathrm{~g}\) The above observations illustrate; |
(a) law of definite proportions.; |
(b) law of conservation of mass; |
(c) law of multiple proportions; |
(d) None of these; |
Q 5. The modern atomic weight scale is based on; |
(a) \(\mathrm{O}^{16}\); |
(b) \(\mathrm{C}^{12}\); |
(c) \(\mathrm{H}^{1}\); |
(d) \(\mathrm{C}^{13}\); |
Q 6. Two samples of lead oxide were separately reduced to metallic lead by heating in a current of hydrogen The weight of lead from one oxide was half the weight of lead obtained from the other oxide The data illustrates; |
(a) law of reciprocal proportions; |
(b) law of constant proportions; |
(c) law of multiple proportions; |
(d) law of equivalent proportions; |
Q 7. How many moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) would be in \(50 \mathrm{~g}\) of the substance ?; |
(a) 0.083 mole; |
(b) 0.952 mole; |
(c) 0.481 mole; |
(d) 0.140 mole; |
Q 8. What is the weight of oxygen required for the complete combustion of \(2 8 \mathrm{~kg}\) of ethylene?; |
(a) \(2.8 \mathrm{~kg}\); |
(b) \(6.4 \mathrm{~kg}\); |
(c) \(9.6 \mathrm{~kg}\); |
(d) \(96 \mathrm{~kg}\); |
Q 9. If 1 5 moles of oxygen combines with \(\mathrm{Al}\) to form \(\mathrm{Al}_{2} \mathrm{O}_{3}\), the mass of \(\mathrm{Al}\) in \(\mathrm{g}\) [Atomic mass of \(\mathrm{Al}=27\) ] used in the reaction is; |
(a) 2.7; |
(b) 54; |
(c) 40.5; |
(d) 81; |
Q 10. The maximum number of molecules are present in; |
(a) \(15 \mathrm{~L}^{2} \mathrm{H}_{2}\) gas at STP; |
(b) \(5 \mathrm{~L}\) of \(\mathrm{N}_{2}\) gas at STP; |
(c) \(0.5 \mathrm{~g}\) of \(\mathrm{H}_{2}\) gas; |
(d) \(10 \mathrm{~g}\) of \(\mathrm{O}_{2}\) gas; |
Classification of Matter
Definition: Matter can be classified based on its physical and chemical properties.
Physical Classification:
Chemical Classification:
Applications:
Q 1. The prefix \(10^{-24}\) is; |
(a) yotta; |
(b) zeta; |
(c) yocto; |
(d) zepto; |
Q 2. When \(25 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is dissolved in \(10^{3} \mathrm{Kg}\) of solution, its concentration will be; |
(A) \(2.5 \mathrm{ppm}\); |
(B) \(25 \mathrm{ppm}\); |
(C) \(250 \mathrm{ppm}\); |
(D) \(100 \mathrm{ppm}\); |
Q 3. In the final answer of the expression \[ \frac{(29.2-20.2)\left(1.79 \times 10^{5}\right)}{1.37} \] the number of significant figures is :; |
(a) 1; |
(b) 2; |
(c) 3; |
(d) 4; |
Q 4. In one experiment, \(4 \mathrm{~g}\) of \(\mathrm{H}_{2}\) combine with \(32 \mathrm{~g}\) of \(\mathrm{O}_{2}\) to form \(36 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) In another experiment, when \(50 \mathrm{~g}_{\mathrm{of}} \mathrm{H}_{2}\) combine with \(400 \mathrm{~g}\) of \(\mathrm{O}_{2}\) then \(450 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) is formed Above two experiments follow; |
(a) The law of conservation of mass; |
(b) The law of constant composition; |
(c) Both (a) and (b); |
(d) Neither (a) nor (b); |
Q 5. The percentage weight of \(\mathrm{Zn}\) in white vitriol \(\left[\mathrm{ZnSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\right]\) is approximately equal to \((\mathrm{Zn}=65, \mathrm{~S}=32, \mathrm{O}=16\) and \(\mathrm{H}=1\) ); |
(a) \(33.65 \%\); |
(b) \(32.56 \%\); |
(c) \(23.65 \%\); |
(d) \(22.65 \%\); |
Q 6. In compound \(\mathrm{A}, 1 00 \mathrm{~g}\) of nitrogen unites with \(0 57 \mathrm{~g}\) of oxygen In compound B, \(2 00 \mathrm{~g}\) of nitrogen combines with \(2 24 \mathrm{~g}\) of oxygen In compound \(\mathrm{C}, 3 00 \mathrm{~g}\) of nitrogen combines with \(5 11 \mathrm{~g}\) of oxygen These results obey the following law; |
(a) law of constant proportion; |
(b) law of multiple proportion; |
(c) law of reciprocal proportion; |
(d) Dalton's law of partial pressure; |
Q 7. The mass of 1 mole of electrons is; |
(a) \(9.1 \times 10^{-28} \mathrm{~g}\); |
(b) \(1.008 \mathrm{mg}\); |
(c) \(0.55 \mathrm{mg}\); |
(d) \(9.1 \times 10^{-27} \mathrm{~g}\); |
Q 8. In the reaction \[ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \] When 1 mole of ammonia and \(1 \mathrm{~mole} \mathrm{of}_{2}\) are made to react to completion,; |
(a) 1.0 mole of \(\mathrm{H}_{2} \mathrm{O}\) is produced; |
(b) 1.0 mole of \(\mathrm{NO}\) will be produced; |
(c) all the oxygen will be consumed; |
(d) all the ammonia will be consumed; |
Q 9. Which one of the following is the lightest?; |
(a) 0.2 mole of hydrogen gas; |
(b) \(6.023 \times 10^{22}\) molecules of nitrogen; |
(c) \(0.1 \mathrm{~g}\) of silver; |
(d) 0.1 mole of oxygen gas; |
Q 10. How many moles of magnesium phosphate, \(\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) will contain 0 25 mole of oxygen atoms?; |
(a) \(1.25 \times 10^{-2}\); |
(b) \(2.5 \times 10^{-2}\); |
(c) 0.02; |
(d) \(3.125 \times 10^{-2}\); |
Laws of Chemical Combination |
Atomic and Molecular Masses |
Mole Concept and Molar Masses |
Percentage Composition |
Stoichiometry and Stoichiometric Calculations |
Concentration of Solutions |
Limiting Reagent |
Empirical and Molecular Formulas |
Avogadro's Law |
Gay-Lussac's Law |
Law of Definite Proportions |
Balancing Chemical Equations |
States of Matter |
Nature of Matter |
Classification of Matter |