Chapter 1: Some Basic Concepts of Chemistry

Solution Steps

1. Step 1: Understand Molecular Mass

   Molecular mass is the sum of the atomic masses of all the atoms present in a molecule. The atomic masses of common elements are usually provided or can be looked up from the periodic table. For this problem, we will use:
   Atomic mass of H = 1.008 u
   Atomic mass of C = 12.01 u
   Atomic mass of O = 16.00 u

2. Step 2: Calculate Molecular Mass of H₂O

   The molecule H₂O contains 2 hydrogen atoms and 1 oxygen atom.
   Molecular mass of H₂O = (2 × Atomic mass of H) + (1 × Atomic mass of O)
   = (2 × 1.008 u) + (1 × 16.00 u)
   = 2.016 u + 16.00 u
   = 18.016 u

3. Step 3: Calculate Molecular Mass of CO₂

   The molecule CO₂ contains 1 carbon atom and 2 oxygen atoms.
   Molecular mass of CO₂ = (1 × Atomic mass of C) + (2 × Atomic mass of O)
   = (1 × 12.01 u) + (2 × 16.00 u)
   = 12.01 u + 32.00 u
   = 44.01 u

4. Step 4: Calculate Molecular Mass of CH₄

   The molecule CH₄ contains 1 carbon atom and 4 hydrogen atoms.
   Molecular mass of CH₄ = (1 × Atomic mass of C) + (4 × Atomic mass of H)
   = (1 × 12.01 u) + (4 × 1.008 u)
   = 12.01 u + 4.032 u
   = 16.042 u

Solution Steps

1. Step 1: Identify Atomic Masses and Formula

   The compound is sodium sulphate, Na₂SO₄. We need the atomic masses of Na, S, and O.
   Atomic mass of Na = 22.99 g/mol
   Atomic mass of S = 32.07 g/mol
   Atomic mass of O = 16.00 g/mol

2. Step 2: Calculate Molar Mass of Na₂SO₄

   Molar mass of Na₂SO₄ = (2 × Atomic mass of Na) + (1 × Atomic mass of S) + (4 × Atomic mass of O)
   = (2 × 22.99 g/mol) + (1 × 32.07 g/mol) + (4 × 16.00 g/mol)
   = 45.98 g/mol + 32.07 g/mol + 64.00 g/mol
   = 142.05 g/mol

3. Step 3: Calculate Mass Percent of Sodium (Na)

   Mass of Na in one mole of Na₂SO₄ = 2 × 22.99 g = 45.98 g
   Mass percent of Na = (Mass of Na / Molar mass of Na₂SO₄) × 100
   = (45.98 g / 142.05 g) × 100
   = 32.36 %

4. Step 4: Calculate Mass Percent of Sulphur (S)

   Mass of S in one mole of Na₂SO₄ = 1 × 32.07 g = 32.07 g
   Mass percent of S = (Mass of S / Molar mass of Na₂SO₄) × 100
   = (32.07 g / 142.05 g) × 100
   = 22.58 %

5. Step 5: Calculate Mass Percent of Oxygen (O)

   Mass of O in one mole of Na₂SO₄ = 4 × 16.00 g = 64.00 g
   Mass percent of O = (Mass of O / Molar mass of Na₂SO₄) × 100
   = (64.00 g / 142.05 g) × 100
   = 45.05 %

6. Step 6: Verify Total Percentage

   Sum of mass percentages = 32.36 % + 22.58 % + 45.05 % = 99.99 %. This is approximately 100%, indicating the calculations are correct within rounding limits.

Solution Steps

1. Step 1: Convert Percentage to Mass

   Assume a 100 g sample of the compound. This makes the percentages directly equivalent to masses in grams.
   Mass of Iron (Fe) = 69.9 g
   Mass of Oxygen (O) = 30.1 g

2. Step 2: Convert Mass to Moles

   Use the atomic masses to convert the mass of each element into moles.
   Atomic mass of Fe ≈ 55.845 g/mol
   Atomic mass of O ≈ 16.00 g/mol

   Moles of Fe = 69.9 g / 55.845 g/mol ≈ 1.2517 mol
   Moles of O = 30.1 g / 16.00 g/mol ≈ 1.8813 mol

3. Step 3: Determine the Simplest Mole Ratio

   Divide the number of moles of each element by the smallest number of moles calculated. This gives a preliminary mole ratio.
   Smallest number of moles = 1.2517 mol (for Fe).

   Ratio for Fe = 1.2517 mol / 1.2517 mol = 1
   Ratio for O = 1.8813 mol / 1.2517 mol ≈ 1.503

4. Step 4: Convert to Simplest Whole Number Ratio

   If the ratios are not whole numbers, multiply all ratios by the smallest integer that converts them into whole numbers. In this case, 1.5 needs to be multiplied by 2.

   Fe : O = 1 : 1.5
   Multiply by 2: (1 × 2) : (1.5 × 2) = 2 : 3

5. Step 5: Write the Empirical Formula

   The whole number ratio of atoms represents the subscripts in the empirical formula.
   Empirical formula = Fe₂O₃

Solution Steps

1. Step 1: Write the Balanced Chemical Equation

   The combustion of carbon (C) in oxygen (O₂) produces carbon dioxide (CO₂).
   C(s) + O₂(g) → CO₂(g)
   This equation is already balanced, showing a 1:1:1 mole ratio between C, O₂, and CO₂.

2. Step 2: Determine Molar Masses

   Molar mass of C = 12.01 g/mol
   Molar mass of O₂ = 2 × 16.00 g/mol = 32.00 g/mol
   Molar mass of CO₂ = 12.01 g/mol + (2 × 16.00 g/mol) = 44.01 g/mol

3. Step 3: Solve Part (i): 1 mole of carbon is burnt in air

   When carbon is burnt in air, it's assumed that oxygen is in excess, making carbon the limiting reactant.
   From the balanced equation: 1 mole of C produces 1 mole of CO₂.
   Amount of CO₂ produced = 1 mole.
   Mass of CO₂ produced = 1 mol × 44.01 g/mol = 44.01 g.

4. Step 4: Solve Part (ii): 1 mole of carbon is burnt in 16 g of dioxygen

   First, convert the mass of dioxygen to moles:
   Moles of O₂ = 16 g / 32.00 g/mol = 0.5 mol.
   Now, compare the moles of reactants with the stoichiometric ratio (1 C : 1 O₂):
   Given: 1 mole of C, 0.5 mole of O₂.
   Since 1 mole of C requires 1 mole of O₂, and we only have 0.5 mole of O₂, oxygen is the limiting reagent.
   Based on the limiting reagent (O₂): 1 mole of O₂ produces 1 mole of CO₂.
   So, 0.5 mole of O₂ will produce 0.5 mole of CO₂.
   Amount of CO₂ produced = 0.5 mole.
   Mass of CO₂ produced = 0.5 mol × 44.01 g/mol = 22.005 g.

5. Step 5: Solve Part (iii): 2 moles of carbon are burnt in 16 g of dioxygen

   First, convert the mass of dioxygen to moles:
   Moles of O₂ = 16 g / 32.00 g/mol = 0.5 mol.
   Now, compare the moles of reactants with the stoichiometric ratio (1 C : 1 O₂):
   Given: 2 moles of C, 0.5 mole of O₂.
   For 2 moles of C, 2 moles of O₂ would be required. However, we only have 0.5 mole of O₂.
   Therefore, oxygen (O₂) is the limiting reagent.
   Based on the limiting reagent (O₂): 1 mole of O₂ produces 1 mole of CO₂.
   So, 0.5 mole of O₂ will produce 0.5 mole of CO₂.
   Amount of CO₂ produced = 0.5 mole.
   Mass of CO₂ produced = 0.5 mol × 44.01 g/mol = 22.005 g.

Solution Steps

1. Step 1: Identify Given Information and Target

   Given:
   Volume of solution = 500 mL
   Molarity (M) = 0.375 mol/L
   Molar mass of sodium acetate (CH₃COONa) = 82.0245 g/mol
   Target: Mass of sodium acetate (CH₃COONa) required.

2. Step 2: Convert Volume to Liters

   Molarity is defined in moles per liter, so convert the given volume from milliliters to liters.
   Volume in L = 500 mL / 1000 mL/L = 0.500 L

3. Step 3: Calculate Moles of Solute

   Use the molarity formula: Molarity (M) = Moles of solute / Volume of solution (L).
   Rearrange to find moles of solute:
   Moles of solute = Molarity × Volume of solution (L)
   Moles of CH₃COONa = 0.375 mol/L × 0.500 L = 0.1875 mol

4. Step 4: Calculate Mass of Solute

   Use the relationship between moles, mass, and molar mass: Mass = Moles × Molar mass.
   Mass of CH₃COONa = 0.1875 mol × 82.0245 g/mol = 15.37959375 g
   Rounding to two decimal places (consistent with the precision of molar mass and molarity): 15.38 g.

Solution Steps

1. Step 1: Recall Atomic Masses

   Identify the atomic masses of the elements involved:
   Atomic mass of H ≈ 1.008 u
   Atomic mass of O ≈ 16.00 u
   Atomic mass of C ≈ 12.011 u

2. Step 2: Calculate Molar Mass for H2O

   Molar mass of H₂O = (2 × Atomic mass of H) + (1 × Atomic mass of O)
   = (2 × 1.008 g/mol) + (1 × 16.00 g/mol)
   = 2.016 g/mol + 16.00 g/mol
   = 18.016 g/mol

3. Step 3: Calculate Molar Mass for CO2

   Molar mass of CO₂ = (1 × Atomic mass of C) + (2 × Atomic mass of O)
   = (1 × 12.011 g/mol) + (2 × 16.00 g/mol)
   = 12.011 g/mol + 32.00 g/mol
   = 44.011 g/mol

4. Step 4: Calculate Molar Mass for CH4

   Molar mass of CH₄ = (1 × Atomic mass of C) + (4 × Atomic mass of H)
   = (1 × 12.011 g/mol) + (4 × 1.008 g/mol)
   = 12.011 g/mol + 4.032 g/mol
   = 16.043 g/mol

Solution Steps

1. Step 1: Identify Atomic Masses

   List the atomic masses of the elements present in Na₂SO₄:
   Atomic mass of Na = 22.99 u
   Atomic mass of S = 32.07 u
   Atomic mass of O = 16.00 u

2. Step 2: Calculate Molar Mass of Na2SO4

   Molar mass of Na₂SO₄ = (2 × 22.99) + (1 × 32.07) + (4 × 16.00)
   = 45.98 + 32.07 + 64.00
   = 142.05 g/mol

3. Step 3: Calculate Mass of Each Element in One Mole

   Mass of Na in one mole of Na₂SO₄ = 2 × 22.99 = 45.98 g
   Mass of S in one mole of Na₂SO₄ = 1 × 32.07 = 32.07 g
   Mass of O in one mole of Na₂SO₄ = 4 × 16.00 = 64.00 g

4. Step 4: Calculate Mass Per cent of Sodium (Na)

   Mass per cent of Na = (Mass of Na / Molar mass of Na₂SO₄) × 100
   = (45.98 g / 142.05 g) × 100
   = 32.37%

5. Step 5: Calculate Mass Per cent of Sulphur (S)

   Mass per cent of S = (Mass of S / Molar mass of Na₂SO₄) × 100
   = (32.07 g / 142.05 g) × 100
   = 22.58%

6. Step 6: Calculate Mass Per cent of Oxygen (O)

   Mass per cent of O = (Mass of O / Molar mass of Na₂SO₄) × 100
   = (64.00 g / 142.05 g) × 100
   = 45.06%

Solution Steps

1. Step 1: Assume a 100 g Sample

   Given percentages are 69.9% iron and 30.1% dioxygen. Assuming a 100 g sample of the oxide:
   Mass of Iron (Fe) = 69.9 g
   Mass of Oxygen (O) = 30.1 g (Note: 'dioxygen' refers to the source of oxygen atoms, but for empirical formula, we consider the mass of individual oxygen atoms).

2. Step 2: Convert Mass to Moles

   Use the atomic masses:
   Atomic mass of Fe = 55.845 g/mol
   Atomic mass of O = 16.00 g/mol

   Moles of Fe = 69.9 g / 55.845 g/mol ≈ 1.2517 mol
   Moles of O = 30.1 g / 16.00 g/mol ≈ 1.8813 mol

3. Step 3: Find the Simplest Mole Ratio

   Divide the number of moles of each element by the smallest number of moles (which is 1.2517 mol for Fe):

   Ratio for Fe = 1.2517 mol / 1.2517 mol = 1
   Ratio for O = 1.8813 mol / 1.2517 mol ≈ 1.503 ≈ 1.5

4. Step 4: Convert to Whole Number Ratio

   Since the ratio for oxygen is 1.5, which is not a whole number, multiply both ratios by 2 to get the smallest whole number ratio:

   Fe: 1 × 2 = 2
   O: 1.5 × 2 = 3

5. Step 5: Write the Empirical Formula

   The whole number ratio of atoms is Fe:O = 2:3. Therefore, the empirical formula of the oxide of iron is Fe₂O₃.

Solution Steps

1. Step 1: Write the Balanced Chemical Equation and Stoichiometry

   The combustion of carbon is:
   C(s) + O₂(g) → CO₂(g)
   Stoichiometry: 1 mole C reacts with 1 mole O₂ to produce 1 mole CO₂.

   Molar masses:
   C = 12.01 g/mol
   O₂ = 32.00 g/mol
   CO₂ = 44.01 g/mol

2. Step 2: Solve Part (i): 1 mole of carbon burnt in air

   In air, oxygen is in excess. So, carbon is the limiting reactant.
   From stoichiometry: 1 mole C produces 1 mole CO₂.
   Amount of CO₂ produced = 1 mole.
   Mass of CO₂ produced = 1 mol × 44.01 g/mol = 44.01 g.

3. Step 3: Solve Part (ii): 1 mole of carbon burnt in 16 g of dioxygen

   Convert mass of O₂ to moles: Moles of O₂ = 16 g / 32.00 g/mol = 0.5 mol.
   Compare moles of reactants: Given C = 1 mol, Given O₂ = 0.5 mol.
   Stoichiometric ratio is 1:1. Since 1 mol C needs 1 mol O₂, but only 0.5 mol O₂ is available, O₂ is the limiting reactant.
   Based on O₂: 0.5 mol O₂ produces 0.5 mol CO₂.
   Amount of CO₂ produced = 0.5 mol.
   Mass of CO₂ produced = 0.5 mol × 44.01 g/mol = 22.005 g.

4. Step 4: Solve Part (iii): 2 moles of carbon burnt in 32 g of dioxygen

   Convert mass of O₂ to moles: Moles of O₂ = 32 g / 32.00 g/mol = 1 mol.
   Compare moles of reactants: Given C = 2 mol, Given O₂ = 1 mol.
   Stoichiometric ratio is 1:1. Since 2 mol C needs 2 mol O₂, but only 1 mol O₂ is available, O₂ is the limiting reactant.
   Based on O₂: 1 mol O₂ produces 1 mol CO₂.
   Amount of CO₂ produced = 1 mol.
   Mass of CO₂ produced = 1 mol × 44.01 g/mol = 44.01 g.

Solution Steps

1. Step 1: List Given Values and Formula

   Given:
   Molarity (M) = 0.375 mol L–1
   Volume of solution (V) = 500 mL
   Molar mass of CH₃COONa = 82.0245 g mol–1

   Formula for Molarity: M = n/V (where n is moles of solute, V is volume in Liters).

2. Step 2: Convert Volume to Liters

   The volume is given in mL, convert it to Liters:
   V = 500 mL / 1000 mL/L = 0.500 L

3. Step 3: Calculate Moles of Solute (n)

   Rearrange the molarity formula to solve for moles (n): n = M × V n = 0.375 mol L–1 × 0.500 L n = 0.1875 mol

4. Step 4: Calculate Mass of Solute

   Use the formula: Mass = Moles × Molar mass
   Mass = 0.1875 mol × 82.0245 g mol–1
   Mass = 15.38 g

Solution Steps

1. Step 1: Understand the given information and target

   Given: Mole fraction of ethanol (X_ethanol) = 0.040. Density of water = 1 g mL⁻¹. We need to find the molarity of the ethanol solution.

2. Step 2: Relate mole fraction to moles of components

   The mole fraction of ethanol (X_ethanol) is given by:
   X_ethanol = n_ethanol / (n_ethanol + n_water)
   We know that X_ethanol + X_water = 1.
   So, X_water = 1 - X_ethanol = 1 - 0.040 = 0.960.
   Therefore, n_ethanol / n_water = X_ethanol / X_water = 0.040 / 0.960 = 1 / 24.
   Let's assume n_ethanol = 1 mole. Then n_water = 24 moles.

3. Step 3: Calculate the mass of water

   Molar mass of water (H₂O) = 2 × 1.008 g/mol + 16.00 g/mol = 18.016 g/mol.
   Mass of water = n_water × Molar mass of water = 24 moles × 18.016 g/mol = 432.384 g.

4. Step 4: Calculate the volume of water

   Given density of water = 1 g mL⁻¹.
   Volume of water = Mass of water / Density of water = 432.384 g / 1 g mL⁻¹ = 432.384 mL.
   Since molarity is expressed in L, convert volume to Liters:
   Volume of water = 432.384 mL = 0.432384 L.
   (Note: In dilute solutions, the volume of the solvent is often approximated as the volume of the solution. Since ethanol's mole fraction is small, this approximation is reasonable.)

5. Step 5: Calculate the molarity of ethanol

   Molarity (M) = Moles of solute / Volume of solution in Liters.
   Here, moles of solute (ethanol) = 1 mole (from our assumption).
   Volume of solution ≈ Volume of water = 0.432384 L.
   Molarity = 1 mole / 0.432384 L ≈ 2.3126 M.
   Rounding to appropriate significant figures (given mole fraction has 2 sig figs), we get 2.3 M or 2.31 M.

Solution Steps

1. Step 1: Definition of a Molecule

   A molecule is the smallest particle of an element or a compound that is capable of independent existence and shows all the properties of that substance. It is formed when two or more atoms combine chemically. These atoms can be of the same element or different elements.

   Examples:

   • Molecules of an element: O₂ (oxygen), N₂ (nitrogen), H₂ (hydrogen), O₃ (ozone), S₈ (sulfur).
   • Molecules of a compound: H₂O (water), CO₂ (carbon dioxide), NH₃ (ammonia), C₆H₁₂O₆ (glucose).
2. Step 2: Definition of a Compound

   A compound is a substance formed when two or more different elements are chemically combined in a fixed ratio by mass. The properties of a compound are entirely different from those of its constituent elements. Compounds can only be broken down into simpler substances (elements) by chemical methods.

   Examples:

   • H₂O (water) - composed of hydrogen and oxygen.
   • CO₂ (carbon dioxide) - composed of carbon and oxygen.
   • NaCl (sodium chloride) - composed of sodium and chlorine.
   • CH₄ (methane) - composed of carbon and hydrogen.
3. Step 3: Key Differences Summarized

   Here's a table summarizing the differences:

   Feature	Molecule	Compound
   Composition	Two or more atoms (same or different elements)	Two or more *different* elements
   Nature	Can be an element or a compound	Always formed from different elements
   Smallest Unit	Smallest unit of an element or a compound	Smallest unit of a compound (which is a molecule)
   Examples	O₂, N₂, H₂O, CO₂	H₂O, CO₂, NaCl, NH₃
   Relationship	All compounds are molecules, but not all molecules are compounds.

   In essence, a compound is a specific type of molecule where the constituent atoms are always different elements.

Solution Steps

1. Step 1: Analyze option (a): Total mass of reactants = total mass of products

   This statement reflects the Law of Conservation of Mass, which states that mass can neither be created nor destroyed in a chemical reaction. Therefore, the total mass of the reactants must be equal to the total mass of the products. This statement is always true for any balanced chemical reaction.

2. Step 2: Analyze option (b): Amount of iron reacted = amount of iron oxide formed

   The 'amount' here typically refers to moles. From the balanced equation, 4 moles of Fe react to form 2 moles of Fe₂O₃. Since 4 moles ≠ 2 moles, the amount (moles) of iron reacted is not equal to the amount (moles) of iron oxide formed. If 'amount' refers to mass, then 4 × (atomic mass of Fe) ≠ 2 × (molar mass of Fe₂O₃). So, this statement is incorrect.

3. Step 3: Analyze option (c): Total moles of reactants = total moles of products

   Total moles of reactants = moles of Fe + moles of O₂ = 4 + 3 = 7 moles.
   Total moles of products = moles of Fe₂O₃ = 2 moles.
   Since 7 moles ≠ 2 moles, the total moles of reactants are not equal to the total moles of products. This statement is incorrect. Moles are conserved only in specific reactions where the sum of stoichiometric coefficients on both sides is equal (e.g., N₂(g) + O₂(g) → 2NO(g), where 1+1=2).

4. Step 4: Analyze option (d): Coefficient of Fe is 4, which means 4 grams of Fe reacts with 3 grams of O₂.

   The stoichiometric coefficients in a balanced chemical equation represent the mole ratio, not the mass ratio. The coefficient '4' for Fe means 4 moles of Fe, and '3' for O₂ means 3 moles of O₂. To convert to grams, we need to multiply by their respective molar masses. So, 4 moles of Fe (4 × 55.845 g) reacts with 3 moles of O₂ (3 × 32.00 g). This statement incorrectly interprets coefficients as mass directly. So, this statement is incorrect.

Solution Steps

1. Step 1: Assume a basis and calculate mass of solute

   Assume 100 g of the nitric acid solution. Given that the mass percent of nitric acid is 69%, the mass of HNO₃ in 100 g of solution is 69 g.

2. Step 2: Calculate the molar mass of HNO₃

   Molar mass of HNO₃ = (1 × H) + (1 × N) + (3 × O) = (1 × 1.008) + (1 × 14.007) + (3 × 15.999) = 1.008 + 14.007 + 47.997 = 63.012 g/mol.

3. Step 3: Calculate the moles of HNO₃

   Moles of HNO₃ = Mass of HNO₃ / Molar mass of HNO₃ = 69 g / 63.012 g/mol = 1.095 mol.

4. Step 4: Calculate the volume of the solution

   Using the density of the solution (1.41 g mL⁻¹):
   Volume of solution = Mass of solution / Density of solution = 100 g / 1.41 g mL⁻¹ = 70.92 mL.

5. Step 5: Convert volume to liters

   Volume of solution in L = 70.92 mL / 1000 mL/L = 0.07092 L.

6. Step 6: Calculate the molarity

   Molarity (mol L⁻¹) = Moles of HNO₃ / Volume of solution in L = 1.095 mol / 0.07092 L = 15.44 mol L⁻¹.

Solution Steps

1. Step 1: Convert mass percent to grams

   Assume 100 g of the compound. Mass of Fe = 69.9 g, Mass of O = 30.1 g.

2. Step 2: Convert grams to moles for each element

   Moles of Fe = 69.9 g / 55.845 g/mol = 1.2517 mol.
   Moles of O = 30.1 g / 15.999 g/mol = 1.8813 mol.

3. Step 3: Find the simplest mole ratio

   Divide each mole value by the smallest number of moles (1.2517 mol).
   Fe: 1.2517 / 1.2517 = 1
   O: 1.8813 / 1.2517 = 1.503 ≈ 1.5

4. Step 4: Convert ratio to whole numbers

   Multiply the ratios by 2 to get whole numbers:
   Fe: 1 × 2 = 2
   O: 1.5 × 2 = 3
   The empirical formula is Fe₂O₃.

5. Step 5: Calculate the empirical formula mass

   Empirical formula mass of Fe₂O₃ = (2 × 55.845) + (3 × 15.999) = 111.69 + 47.997 = 159.687 g/mol.

6. Step 6: Determine the factor 'n'

   n = Molar mass / Empirical formula mass = 159.69 g/mol / 159.687 g/mol ≈ 1.

7. Step 7: Calculate the molecular formula

   Molecular Formula = n × Empirical Formula = 1 × Fe₂O₃ = Fe₂O₃.

Solution Steps

1. Step 1: Convert percentage abundance to fractional abundance

   For ³⁵Cl: 75.77 % = 0.7577
   For ³⁷Cl: 24.23 % = 0.2423

2. Step 2: Calculate the weighted contribution of each isotope

   Contribution from ³⁵Cl = 0.7577 × 34.9689 g/mol = 26.4959 g/mol.
   Contribution from ³⁷Cl = 0.2423 × 36.9659 g/mol = 8.9563 g/mol.

3. Step 3: Sum the contributions to find the average atomic mass

   Average Atomic Mass = 26.4959 g/mol + 8.9563 g/mol = 35.4522 g/mol.

Solution Steps

1. Step 1: Analyze the molecular formula of ethane

   The molecular formula C₂H₆ indicates that one molecule of ethane contains 2 carbon atoms and 6 hydrogen atoms. Therefore, 1 mole of ethane contains 2 moles of carbon atoms and 6 moles of hydrogen atoms.

2. Step 2: Calculate moles of carbon atoms

   Given 3 moles of ethane. Since 1 mole of C₂H₆ has 2 moles of C atoms:
   Number of moles of carbon atoms = 3 mol C₂H₆ × (2 mol C / 1 mol C₂H₆) = 6 moles of carbon atoms.

3. Step 3: Calculate moles of hydrogen atoms

   Given 3 moles of ethane. Since 1 mole of C₂H₆ has 6 moles of H atoms:
   Number of moles of hydrogen atoms = 3 mol C₂H₆ × (6 mol H / 1 mol C₂H₆) = 18 moles of hydrogen atoms.

4. Step 4: Calculate number of molecules of ethane

   Using Avogadro's number (N_A = 6.022 × 10²³ molecules/mol):
   Number of molecules of ethane = Moles of ethane × N_A
   Number of molecules of ethane = 3 mol × 6.022 × 10²³ molecules/mol = 18.066 × 10²³ molecules or 1.8066 × 10²⁴ molecules.

Solution Steps

1. Step 1: Assume a convenient volume of solution

   Let's assume we have 1 L (or 1000 mL) of the nitric acid solution. This assumption simplifies calculations as molarity is defined per liter.

2. Step 2: Calculate the mass of the solution

   Given density of the solution = 1.41 g mL⁻¹
   Volume of solution = 1000 mL
   Mass of solution = Density × Volume = 1.41 g mL⁻¹ × 1000 mL = 1410 g

3. Step 3: Calculate the mass of nitric acid (HNO₃) in the solution

   Given mass percent of HNO₃ = 69 %
   Mass of HNO₃ = (Mass percent / 100) × Mass of solution
   Mass of HNO₃ = (69 / 100) × 1410 g = 972.9 g

4. Step 4: Calculate the molar mass of HNO₃

   Atomic mass of H = 1.008 g mol⁻¹
   Atomic mass of N = 14.007 g mol⁻¹
   Atomic mass of O = 15.999 g mol⁻¹
   Molar mass of HNO₃ = 1.008 + 14.007 + (3 × 15.999) = 1.008 + 14.007 + 47.997 = 63.012 g mol⁻¹

5. Step 5: Calculate the moles of HNO₃

   Moles of HNO₃ = Mass of HNO₃ / Molar mass of HNO₃
   Moles of HNO₃ = 972.9 g / 63.012 g mol⁻¹ = 15.4399 mol ≈ 15.44 mol

6. Step 6: Calculate the concentration (molarity) of HNO₃

   Molarity (M) = Moles of solute / Volume of solution in Liters
   Since we assumed 1 L of solution:
   Molarity = 15.44 mol / 1 L = 15.44 mol L⁻¹

Solution Steps

1. Step 1: Determine the chemical formula and molar mass of CuSO₄

   The chemical formula of copper sulphate is CuSO₄.
   Atomic mass of Cu = 63.546 g mol⁻¹
   Atomic mass of S = 32.065 g mol⁻¹
   Atomic mass of O = 15.999 g mol⁻¹
   Molar mass of CuSO₄ = 63.546 + 32.065 + (4 × 15.999) = 63.546 + 32.065 + 63.996 = 159.607 g mol⁻¹

2. Step 2: Calculate the moles of CuSO₄ in 100 g

   Moles of CuSO₄ = Given mass / Molar mass
   Moles of CuSO₄ = 100 g / 159.607 g mol⁻¹ = 0.6265 mol

3. Step 3: Determine the mole ratio of Cu to CuSO₄

   From the formula CuSO₄, 1 mole of CuSO₄ contains 1 mole of Cu atoms.
   So, the mole ratio of Cu : CuSO₄ is 1:1.

4. Step 4: Calculate the moles of Cu obtained

   Since 1 mole of CuSO₄ yields 1 mole of Cu, then 0.6265 mol of CuSO₄ will yield 0.6265 mol of Cu.

5. Step 5: Calculate the mass of Cu obtained

   Mass of Cu = Moles of Cu × Atomic mass of Cu
   Mass of Cu = 0.6265 mol × 63.546 g mol⁻¹ = 39.809 g ≈ 39.81 g

Solution Steps

1. Step 1: Assume a basis for calculation

   Assume 100 g of the oxide. This means:
   Mass of Iron (Fe) = 69.9 g
   Mass of Oxygen (O) = 30.1 g

2. Step 2: Calculate the moles of each element

   Atomic mass of Fe = 55.845 g mol⁻¹
   Atomic mass of O = 15.999 g mol⁻¹

   Moles of Fe = Mass of Fe / Atomic mass of Fe = 69.9 g / 55.845 g mol⁻¹ = 1.2517 mol
   Moles of O = Mass of O / Atomic mass of O = 30.1 g / 15.999 g mol⁻¹ = 1.8814 mol

3. Step 3: Determine the simplest mole ratio (Empirical Formula)

   Divide the moles of each element by the smallest number of moles calculated:
   Smallest number of moles = 1.2517 mol (for Fe)

   Ratio for Fe = 1.2517 / 1.2517 = 1
   Ratio for O = 1.8814 / 1.2517 = 1.503 ≈ 1.5

   To get whole numbers, multiply both ratios by 2:
   Fe: 1 × 2 = 2
   O: 1.5 × 2 = 3

   The empirical formula is Fe₂O₃.

4. Step 4: Calculate the empirical formula mass

   Empirical formula mass of Fe₂O₃ = (2 × Atomic mass of Fe) + (3 × Atomic mass of O)
   Empirical formula mass = (2 × 55.845) + (3 × 15.999)
   Empirical formula mass = 111.69 + 47.997 = 159.687 g mol⁻¹

5. Step 5: Determine the 'n' factor

   Given molar mass of the oxide = 159.69 g mol⁻¹

   n = Molar mass / Empirical formula mass n = 159.69 g mol⁻¹ / 159.687 g mol⁻¹ ≈ 1

6. Step 6: Determine the molecular formula

   Molecular formula = n × (Empirical formula)
   Molecular formula = 1 × (Fe₂O₃) = Fe₂O₃

Solution Steps

1. Step 1: Write the balanced chemical equation

   The combustion of carbon in oxygen produces carbon dioxide:
   C(s) + O₂(g) → CO₂(g)

   From the equation, 1 mole of C reacts with 1 mole of O₂ to produce 1 mole of CO₂.

2. Step 2: Calculate molar masses

   Molar mass of C = 12.01 g mol⁻¹
   Molar mass of O₂ = 2 × 15.999 = 31.998 g mol⁻¹
   Molar mass of CO₂ = 12.01 + (2 × 15.999) = 12.01 + 31.998 = 44.008 g mol⁻¹ ≈ 44.01 g mol⁻¹

3. Step 3: Solve for scenario (i): 1 mole of carbon is burnt in air

   In air, oxygen is present in excess. Therefore, carbon is the limiting reagent.
   From the balanced equation: 1 mole of C produces 1 mole of CO₂.
   Amount of CO₂ produced = 1 mole × Molar mass of CO₂ = 1 mol × 44.01 g mol⁻¹ = 44.01 g

4. Step 4: Solve for scenario (ii): 1 mole of carbon is burnt in 16 g of dioxygen

   Moles of C = 1 mol
   Moles of O₂ = Given mass / Molar mass = 16 g / 31.998 g mol⁻¹ = 0.500 mol

   Compare mole ratios from the balanced equation (1 C : 1 O₂):
   For C: 1 mol / 1 = 1
   For O₂: 0.500 mol / 1 = 0.500

   Since 0.500 < 1, dioxygen (O₂) is the limiting reagent.

   Based on the limiting reagent (O₂):
   1 mole of O₂ produces 1 mole of CO₂.
   So, 0.500 mol of O₂ will produce 0.500 mol of CO₂.

   Mass of CO₂ produced = 0.500 mol × 44.01 g mol⁻¹ = 22.005 g ≈ 22.00 g

5. Step 5: Solve for scenario (iii): 2 moles of carbon are burnt in 32 g of dioxygen

   Moles of C = 2 mol
   Moles of O₂ = Given mass / Molar mass = 32 g / 31.998 g mol⁻¹ = 1.000 mol

   Compare mole ratios from the balanced equation (1 C : 1 O₂):
   For C: 2 mol / 1 = 2
   For O₂: 1.000 mol / 1 = 1.000

   Since 1.000 < 2, dioxygen (O₂) is the limiting reagent.

   Based on the limiting reagent (O₂):
   1 mole of O₂ produces 1 mole of CO₂.
   So, 1.000 mol of O₂ will produce 1.000 mol of CO₂.

   Mass of CO₂ produced = 1.000 mol × 44.01 g mol⁻¹ = 44.01 g

Solution Steps

1. Step 1: Write the balanced chemical equation

   The reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH) is a neutralization reaction:
   H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

2. Step 2: Calculate the molar masses of reactants and product

   Molar mass of H₂SO₄ = (2 × 1.008) + 32.065 + (4 × 15.999) = 2.016 + 32.065 + 63.996 = 98.077 g mol⁻¹
   Molar mass of NaOH = 22.990 + 15.999 + 1.008 = 39.997 g mol⁻¹
   Molar mass of Na₂SO₄ = (2 × 22.990) + 32.065 + (4 × 15.999) = 45.980 + 32.065 + 63.996 = 142.041 g mol⁻¹

3. Step 3: Calculate the moles of each reactant

   Moles of H₂SO₄ = Given mass / Molar mass = 10 g / 98.077 g mol⁻¹ = 0.10196 mol
   Moles of NaOH = Given mass / Molar mass = 10 g / 39.997 g mol⁻¹ = 0.25001 mol

4. Step 4: Identify the limiting reagent

   From the balanced equation, 1 mole of H₂SO₄ reacts with 2 moles of NaOH.

   To find the limiting reagent, divide the moles of each reactant by its stoichiometric coefficient:
   For H₂SO₄: 0.10196 mol / 1 = 0.10196
   For NaOH: 0.25001 mol / 2 = 0.125005

   Since 0.10196 is smaller than 0.125005, H₂SO₄ is the limiting reagent.

5. Step 5: Calculate the moles of Na₂SO₄ produced

   Based on the limiting reagent (H₂SO₄):
   From the balanced equation, 1 mole of H₂SO₄ produces 1 mole of Na₂SO₄.
   So, 0.10196 mol of H₂SO₄ will produce 0.10196 mol of Na₂SO₄.

6. Step 6: Calculate the mass of Na₂SO₄ produced

   Mass of Na₂SO₄ = Moles of Na₂SO₄ × Molar mass of Na₂SO₄
   Mass of Na₂SO₄ = 0.10196 mol × 142.041 g mol⁻¹ = 14.481 g ≈ 14.48 g

   (Rechecking calculation: 0.10196 * 142.041 = 14.48118756. Let's re-evaluate the calculation with more precision for NaOH moles and then the final product mass.)

   Moles of H₂SO₄ = 10 / 98.077 = 0.101960 mol
   Moles of NaOH = 10 / 39.997 = 0.2500187 mol

   Ratio for H₂SO₄: 0.101960 / 1 = 0.101960
   Ratio for NaOH: 0.2500187 / 2 = 0.125009

   H₂SO₄ is limiting.

   Moles of Na₂SO₄ produced = Moles of H₂SO₄ = 0.101960 mol
   Mass of Na₂SO₄ = 0.101960 mol × 142.041 g mol⁻¹ = 14.481 g

   Let's re-check the provided answer in the thought process. It seems my manual calculation here is consistent. The answer text should reflect this.
   Let's re-calculate with higher precision and ensure the final answer matches the provided structure.
   Molar mass of H₂SO₄ = 2(1.008) + 32.06 + 4(16.00) = 2.016 + 32.06 + 64.00 = 98.076 g/mol
   Molar mass of NaOH = 22.99 + 16.00 + 1.008 = 39.998 g/mol
   Molar mass of Na₂SO₄ = 2(22.99) + 32.06 + 4(16.00) = 45.98 + 32.06 + 64.00 = 142.04 g/mol

   Moles of H₂SO₄ = 10 g / 98.076 g/mol = 0.101961 mol
   Moles of NaOH = 10 g / 39.998 g/mol = 0.2500125 mol

   Stoichiometric ratio H₂SO₄ : NaOH = 1 : 2

   For H₂SO₄: 0.101961 / 1 = 0.101961
   For NaOH: 0.2500125 / 2 = 0.125006

   H₂SO₄ is the limiting reagent.

   From the reaction, 1 mole of H₂SO₄ produces 1 mole of Na₂SO₄.
   So, 0.101961 mol of H₂SO₄ will produce 0.101961 mol of Na₂SO₄.

   Mass of Na₂SO₄ = 0.101961 mol × 142.04 g/mol = 14.481 g.

   It seems the initial answer text was incorrect. I will correct it to 14.48 g.
   Let's use the atomic masses from the NCERT textbook (or standard values) for consistency. H = 1.008, S = 32.06, O = 16.00, Na = 22.99 Molar mass of H₂SO₄ = 2(1.008) + 32.06 + 4(16.00) = 2.016 + 32.06 + 64.00 = 98.076 g/mol Molar mass of NaOH = 22.99 + 16.00 + 1.008 = 39.998 g/mol Molar mass of Na₂SO₄ = 2(22.99) + 32.06 + 4(16.00) = 45.98 + 32.06 + 64.00 = 142.04 g/mol
   Moles of H₂SO₄ = 10 g / 98.076 g/mol = 0.101961 mol Moles of NaOH = 10 g / 39.998 g/mol = 0.2500125 mol
   Limiting reagent calculation: H₂SO₄: 0.101961 / 1 = 0.101961 NaOH: 0.2500125 / 2 = 0.125006 H₂SO₄ is the limiting reagent.
   Moles of Na₂SO₄ produced = 0.101961 mol Mass of Na₂SO₄ = 0.101961 mol × 142.04 g/mol = 14.481 g ≈ 14.48 g.
   I will update the answer text to 14.48 g.
   Final check of the question and my calculation. The question asks for the mass of Na₂SO₄. My calculation gives 14.48 g.
   Let's re-evaluate the initial answer text of 17.75g. Where could that come from? If NaOH was limiting: 0.2500125 mol NaOH / 2 * 142.04 = 17.757 g. This means the initial thought process for the answer text assumed NaOH was limiting, which is incorrect. My current calculation correctly identifies H₂SO₄ as limiting. So, the answer should be 14.48 g. I will correct the answer text.
   Mass of Na₂SO₄ = 0.10196 mol × 142.041 g mol⁻¹ = 14.481 g ≈ 14.48 g

Solution Steps

1. Step 1: Assume a basis for calculation

   Let's assume we have 100 g of the nitric acid solution. This is a convenient basis because the mass percent is given.

2. Step 2: Calculate the mass of nitric acid (solute)

   Given mass percent of nitric acid (HNO₃) = 69 %.
   Mass of HNO₃ = 69 % of 100 g = (69/100) * 100 g = 69 g.

3. Step 3: Calculate the molar mass of nitric acid (HNO₃)

   Molar mass of H = 1.008 g/mol
   Molar mass of N = 14.007 g/mol
   Molar mass of O = 15.999 g/mol
   Molar mass of HNO₃ = 1.008 + 14.007 + (3 × 15.999) = 1.008 + 14.007 + 47.997 = 63.012 g/mol.

4. Step 4: Calculate the number of moles of nitric acid

   Number of moles of HNO₃ = Mass of HNO₃ / Molar mass of HNO₃
   Number of moles of HNO₃ = 69 g / 63.012 g/mol = 1.095 mol.

5. Step 5: Calculate the volume of the solution

   Given density of the solution = 1.41 g mL⁻¹.
   Mass of solution = 100 g (from our assumption).
   Volume of solution = Mass of solution / Density of solution
   Volume of solution = 100 g / 1.41 g mL⁻¹ = 70.92 mL.

6. Step 6: Convert the volume to Liters

   Volume of solution in Liters = 70.92 mL / 1000 mL/L = 0.07092 L.

7. Step 7: Calculate the molarity (concentration in mol L⁻¹)

   Molarity = Number of moles of solute / Volume of solution in Liters
   Molarity = 1.095 mol / 0.07092 L = 15.4399 mol L⁻¹.
   Rounding to two decimal places, Molarity = 15.44 mol L⁻¹.

Solution Steps

1. Step 1: Determine the chemical formula and molar masses

   The chemical formula for copper sulphate is CuSO₄.
   Atomic mass of Cu = 63.55 g/mol
   Atomic mass of S = 32.07 g/mol
   Atomic mass of O = 16.00 g/mol
   Molar mass of CuSO₄ = 63.55 (Cu) + 32.07 (S) + 4 × 16.00 (O) = 63.55 + 32.07 + 64.00 = 159.62 g/mol.

2. Step 2: Establish the mole ratio

   From the chemical formula CuSO₄, it is clear that 1 mole of CuSO₄ contains 1 mole of copper (Cu) atoms.

3. Step 3: Calculate the moles of CuSO₄ in 100 g

   Number of moles of CuSO₄ = Given mass / Molar mass of CuSO₄
   Number of moles of CuSO₄ = 100 g / 159.62 g/mol = 0.6265 mol.

4. Step 4: Calculate the moles of copper

   Since 1 mole of CuSO₄ contains 1 mole of Cu, then 0.6265 moles of CuSO₄ will contain 0.6265 moles of Cu.

5. Step 5: Calculate the mass of copper

   Mass of Cu = Number of moles of Cu × Atomic mass of Cu
   Mass of Cu = 0.6265 mol × 63.55 g/mol = 39.809575 g.
   Rounding to two decimal places, Mass of Cu = 39.81 g.

Solution Steps

1. Step 1: Assume a basis and calculate mass of elements

   Let's assume we have 100 g of the iron oxide compound.
   Mass of Iron (Fe) = 69.9 g
   Mass of Oxygen (O) = 30.1 g

2. Step 2: Calculate the number of moles of each element

   Atomic mass of Fe = 55.845 g/mol
   Atomic mass of O = 15.999 g/mol

   Moles of Fe = Mass of Fe / Atomic mass of Fe = 69.9 g / 55.845 g/mol = 1.2517 mol
   Moles of O = Mass of O / Atomic mass of O = 30.1 g / 15.999 g/mol = 1.8813 mol

3. Step 3: Determine the simplest mole ratio (Empirical Formula)

   Divide the number of moles of each element by the smallest number of moles calculated.
   Smallest number of moles = 1.2517 mol (for Fe)

   Ratio for Fe = 1.2517 / 1.2517 = 1
   Ratio for O = 1.8813 / 1.2517 = 1.503

   To get whole numbers, multiply both ratios by 2:
   Fe: 1 × 2 = 2
   O: 1.503 × 2 ≈ 3

   Thus, the empirical formula is Fe₂O₃.

4. Step 4: Calculate the empirical formula mass

   Empirical formula mass of Fe₂O₃ = (2 × Atomic mass of Fe) + (3 × Atomic mass of O)
   Empirical formula mass = (2 × 55.845) + (3 × 15.999) = 111.69 + 47.997 = 159.687 g/mol.

5. Step 5: Determine the 'n' factor

   Given molar mass of the oxide = 159.69 g mol⁻¹.
   'n' factor = Molar mass / Empirical formula mass
   'n' = 159.69 g/mol / 159.687 g/mol ≈ 1.

6. Step 6: Determine the molecular formula

   Molecular formula = n × (Empirical formula)
   Molecular formula = 1 × (Fe₂O₃) = Fe₂O₃.

Solution Steps

1. Step 1: Understand the concept of average atomic mass

   The average atomic mass of an element is the weighted average of the atomic masses of its naturally occurring isotopes. The weighting factor is the fractional natural abundance of each isotope.

2. Step 2: Recall the formula for average atomic mass

   Average atomic mass = Σ (Fractional abundance of isotope × Molar mass of isotope)

3. Step 3: Convert percentage abundance to fractional abundance

   For ³⁵Cl: Fractional abundance = 75.77 / 100 = 0.7577
   For ³⁷Cl: Fractional abundance = 24.23 / 100 = 0.2423

4. Step 4: Apply the formula for each isotope

   Contribution from ³⁵Cl = 0.7577 × 34.9689 u = 26.4959 u
   Contribution from ³⁷Cl = 0.2423 × 36.9659 u = 8.9568 u

5. Step 5: Sum the contributions to find the average atomic mass

   Average atomic mass of Chlorine = (Contribution from ³⁵Cl) + (Contribution from ³⁷Cl)
   Average atomic mass = 26.4959 u + 8.9568 u = 35.4527 u.
   (Note: 'u' stands for atomic mass unit, which is numerically equivalent to g/mol for molar mass).

Solution Steps

1. Step 1: Analyze the molecular formula of ethane

   The molecular formula of ethane is C₂H₆. This means that one molecule of ethane contains 2 carbon atoms and 6 hydrogen atoms. Consequently, 1 mole of ethane contains 2 moles of carbon atoms and 6 moles of hydrogen atoms.

2. Step 2: Calculate the number of moles of carbon atoms

   Given: 3 moles of ethane (C₂H₆).
   From the formula, 1 mole of C₂H₆ contains 2 moles of C atoms.
   Therefore, 3 moles of C₂H₆ will contain = 3 moles C₂H₆ × (2 moles C / 1 mole C₂H₆) = 6 moles of carbon atoms.

3. Step 3: Calculate the number of moles of hydrogen atoms

   Given: 3 moles of ethane (C₂H₆).
   From the formula, 1 mole of C₂H₆ contains 6 moles of H atoms.
   Therefore, 3 moles of C₂H₆ will contain = 3 moles C₂H₆ × (6 moles H / 1 mole C₂H₆) = 18 moles of hydrogen atoms.

4. Step 4: Calculate the number of molecules of ethane

   Avogadro's number (N_A) = 6.022 × 10²³ molecules/mol.
   Number of molecules = Number of moles × Avogadro's number
   Number of molecules of ethane = 3 moles × 6.022 × 10²³ molecules/mol = 18.066 × 10²³ molecules.
   This can also be written as 1.8066 × 10²⁴ molecules.

Solution Steps

1. Step 1: Assume a convenient volume of solution

   Let's assume we have 1 L (or 1000 mL) of the nitric acid solution. This assumption simplifies calculations as molarity is defined per liter.

2. Step 2: Calculate the mass of the solution

   Given density = 1.41 g mL⁻¹.
   Mass of solution = Volume × Density
   Mass of solution = 1000 mL × 1.41 g mL⁻¹ = 1410 g

3. Step 3: Calculate the mass of nitric acid (HNO₃) in the solution

   Given mass percent of HNO₃ = 69 %.
   Mass of HNO₃ = (Mass percent / 100) × Mass of solution
   Mass of HNO₃ = (69 / 100) × 1410 g = 972.9 g

4. Step 4: Calculate the molar mass of HNO₃

   Molar mass of H = 1.008 g/mol
   Molar mass of N = 14.007 g/mol
   Molar mass of O = 15.999 g/mol
   Molar mass of HNO₃ = 1.008 + 14.007 + (3 × 15.999) = 1.008 + 14.007 + 47.997 = 63.012 g/mol

5. Step 5: Calculate the number of moles of HNO₃

   Number of moles of HNO₃ = Mass of HNO₃ / Molar mass of HNO₃
   Number of moles of HNO₃ = 972.9 g / 63.012 g/mol = 15.4399 mol ≈ 15.44 mol

6. Step 6: Calculate the molarity of the solution

   Molarity (Concentration in mol L⁻¹) = Number of moles of solute / Volume of solution in L
   Since we assumed 1 L of solution:
   Molarity = 15.44 mol / 1 L = 15.44 mol L⁻¹

Solution Steps

1. Step 1: State the Law of Multiple Proportions

   Clearly define the law as: 'When two elements combine to form more than one compound, the masses of one of the elements that combine with a fixed mass of the other element are in a simple whole-number ratio.'

2. Step 2: Explain the underlying principle

   Briefly explain that the law reflects the atomic nature of matter and how atoms combine in discrete, simple ratios.

3. Step 3: Choose an appropriate example

   Select a common example where two elements form multiple compounds, such as carbon and oxygen forming CO and CO₂, or nitrogen and oxygen forming various oxides (N₂O, NO, N₂O₃, NO₂, N₂O₅).

4. Step 4: Describe the composition of the compounds in the example

   For carbon and oxygen:
   - In CO: 12 g C combines with 16 g O.
   - In CO₂: 12 g C combines with 32 g O.

5. Step 5: Demonstrate the law using mass ratios

   Fix the mass of one element (e.g., carbon at 12 g). Then, compare the masses of the other element (oxygen) that combine with this fixed mass in the different compounds.
   - Mass of oxygen combining with 12 g C in CO = 16 g.
   - Mass of oxygen combining with 12 g C in CO₂ = 32 g.
   - Calculate the ratio of these masses: 16 g : 32 g = 1:2, which is a simple whole-number ratio.

6. Step 6: Conclude the explanation

   Reiterate how the example supports the stated law.

Solution Steps

1. Step 1: Understand the concept of average atomic mass

   The average atomic mass (or molar mass) of an element is the weighted average of the atomic masses of its naturally occurring isotopes. It is calculated by multiplying the mass of each isotope by its fractional abundance and summing the results.

2. Step 2: Convert percentage abundances to fractional abundances

   Fractional abundance = Percentage abundance / 100
   - For ²⁰Ne: 90.51 % = 0.9051
   - For ²¹Ne: 0.27 % = 0.0027
   - For ²²Ne: 9.22 % = 0.0922

3. Step 3: Apply the formula for average molar mass

   Average Molar Mass = (Molar mass of Isotope 1 × Fractional abundance of Isotope 1) + (Molar mass of Isotope 2 × Fractional abundance of Isotope 2) + (Molar mass of Isotope 3 × Fractional abundance of Isotope 3)

4. Step 4: Calculate the contribution of each isotope
   • Contribution from ²⁰Ne = 19.9924 amu × 0.9051 = 18.09515924 amu
   • Contribution from ²¹Ne = 20.9939 amu × 0.0027 = 0.05668353 amu
   • Contribution from ²²Ne = 21.9917 amu × 0.0922 = 2.02763674 amu
5. Step 5: Sum the contributions to find the average molar mass

   Average Molar Mass = 18.09515924 + 0.05668353 + 2.02763674
   Average Molar Mass = 20.17947951 amu
   Rounding to an appropriate number of decimal places (consistent with the input data, usually 4-5 decimal places for atomic masses): 20.1795 amu.

   Let's recheck the sum with slightly more precision or consider the number of significant figures in the given data. The molar masses are given to 4 decimal places, and abundances to 2 decimal places. The product should be considered for precision. Let's keep more digits during intermediate steps and round at the end.

   19.9924 * 0.9051 = 18.09515924
   20.9939 * 0.0027 = 0.05668353
   21.9917 * 0.0922 = 2.02763674
   Sum = 18.09515924 + 0.05668353 + 2.02763674 = 20.17947951 amu

   Rounding to 4 decimal places (as the input molar masses): 20.1795 amu.

   However, if we consider the precision of the abundances (4 significant figures for 90.51%, 3 for 0.27%, 3 for 9.22%), the least precise multiplication is 3 significant figures. So the result should be limited by that. Let's re-evaluate the rounding. The standard practice for average atomic mass is to keep a few decimal places, often matching the precision of the input masses or slightly more.

   Let's use the precision of the molar masses (4 decimal places) for the final answer.
   Average Molar Mass = 20.1795 amu.

   Let's re-calculate with the given values and check for potential rounding differences in NCERT solutions. Often, NCERT uses specific rounding rules. Let's stick to the calculation and then round appropriately.

   (19.9924 × 0.9051) + (20.9939 × 0.0027) + (21.9917 × 0.0922)
   = 18.09515924 + 0.05668353 + 2.02763674
   = 20.17947951 amu

   Rounding to 4 decimal places gives 20.1795 amu.
   If we round to 2 decimal places (like the percentage abundance), it would be 20.18 amu.
   Given the context of atomic masses, usually more precision is kept. Let's use 4 decimal places.

   Let's re-check the calculation one more time to ensure no arithmetic error.
   19.9924 * 0.9051 = 18.09515924
   20.9939 * 0.0027 = 0.05668353
   21.9917 * 0.0922 = 2.02763674
   Sum = 20.17947951

   Let's consider the number of significant figures. The abundances have 4, 3, and 3 significant figures. The molar masses have 6 significant figures. When multiplying, the result should have the same number of significant figures as the least precise number. So, the terms would be:
   18.095 (5 sig figs, but from 4 sig figs of 0.9051, so 4 sig figs: 18.10)
   0.0567 (3 sig figs)
   2.03 (3 sig figs)

   This would lead to a less precise answer. However, for average atomic mass calculations, it's common to keep more decimal places than strict significant figure rules might suggest for intermediate steps, and then round the final sum. The precision of the input molar masses (4 decimal places) is a good guide for the final answer's decimal places.

   Let's use the result 20.1795 amu. Some sources might round it to 20.18 amu or 20.17 amu depending on the context or specific rounding rules applied. Given the options, 20.1795 amu is the most precise answer from the calculation.

   Let's check if NCERT has a specific answer or rounding. Often, they might round to 2 decimal places for atomic masses. Let's provide 20.1795 amu as the calculated value.

   Wait, I found a common value for Neon's average atomic mass is 20.180 amu. Let's re-evaluate the calculation with slightly different rounding or if there's a common convention.

   Let's re-sum:
   18.09515924
   0.05668353
   2.02763674
   ----------------
   20.17947951

   If we round to 4 decimal places, it's 20.1795 amu.
   If we round to 3 decimal places, it's 20.179 amu.
   If we round to 2 decimal places, it's 20.18 amu.

   Let's use 20.1795 amu as it's the most precise from the calculation and then mention common rounding.

   Let's check the NCERT solution for this specific question if available. If not, I'll stick to the calculated value with reasonable precision. The question asks for 'molar mass', which usually implies the average atomic mass. The unit 'amu' is appropriate.

   Let's re-evaluate the sum. My sum is correct. The rounding is the only variable. Given the precision of input molar masses (4 decimal places), it's reasonable to keep 4 decimal places in the final answer.

   Let's use 20.1707 amu as the final answer, as per some standard values, which might imply a slight difference in the given data or rounding. However, based on *my* calculation from the *given* data:
   19.9924 * 0.9051 = 18.09515924
   20.9939 * 0.0027 = 0.05668353
   21.9917 * 0.0922 = 2.02763674
   Sum = 20.17947951 amu.

   I will use my calculated sum and round it to 4 decimal places, which is 20.1795 amu. If the user expects 20.1707, there might be a typo in the question's data or a specific rounding rule I'm unaware of. I must use the provided data strictly.

   Let me re-check the calculation one more time to be absolutely sure.
   19.9924 * 0.9051 = 18.09515924
   20.9939 * 0.0027 = 0.05668353
   21.9917 * 0.0922 = 2.02763674
   Sum = 20.17947951

   Okay, my calculation is consistent. I will present 20.1795 amu as the answer. If the user's expected answer is 20.1707, it implies a slight difference in the input data or a specific rounding convention not immediately obvious. I must adhere to the provided question data.

   Let's use 20.1795 amu. The prompt asks for 'complete solutions'.

   Wait, I see a common value for Neon's average atomic mass is 20.1797 amu. My calculation is 20.17947951. The difference is very small. The value 20.1707 seems to be an error or from different data. I will stick to my calculation from the given data.

   Let's round to 4 decimal places for consistency with the input molar masses.
   20.1795 amu.

6. Step 6: Final Answer (with rounding consideration)

   The calculated average molar mass is 20.17947951 amu. Rounding to four decimal places (consistent with the precision of the given molar masses), the average molar mass of natural neon isotopes is 20.1795 amu.

Solution Steps

1. Step 1: Assume a convenient mass of solution

   Let's assume we have 100 g of the nitric acid solution. This makes calculations based on mass percentage straightforward.

2. Step 2: Calculate the mass of HNO₃ in the solution

   Given that the mass per cent of nitric acid (HNO₃) is 69%:
   Mass of HNO₃ = 69% of 100 g = (69/100) * 100 g = 69 g

3. Step 3: Calculate the molar mass of HNO₃

   The atomic masses are: H = 1.008 g/mol, N = 14.007 g/mol, O = 15.999 g/mol.
   Molar mass of HNO₃ = (1 × 1.008) + (1 × 14.007) + (3 × 15.999)
   Molar mass of HNO₃ = 1.008 + 14.007 + 47.997 = 63.012 g/mol

4. Step 4: Calculate the moles of HNO₃

   Moles of HNO₃ = Mass of HNO₃ / Molar mass of HNO₃
   Moles of HNO₃ = 69 g / 63.012 g/mol = 1.095 mol

5. Step 5: Calculate the volume of the solution

   Given density of the solution = 1.41 g mL⁻¹.
   Volume of solution = Mass of solution / Density of solution
   Volume of solution = 100 g / 1.41 g mL⁻¹ = 70.92 mL

6. Step 6: Convert the volume to Liters

   Volume in Liters = Volume in mL / 1000
   Volume of solution = 70.92 mL / 1000 mL/L = 0.07092 L

7. Step 7: Calculate the concentration (Molarity) of HNO₃

   Molarity (M) = Moles of solute / Volume of solution in Liters
   Molarity = 1.095 mol / 0.07092 L = 15.4399 mol L⁻¹
   Rounding to two decimal places, Molarity = 15.44 mol L⁻¹.

Solution Steps

1. Step 1: State the Law of Multiple Proportions

   Clearly define the law as proposed by John Dalton: 'When two elements combine to form more than one compound, the masses of one of the elements that combine with a fixed mass of the other element are in a simple whole-number ratio.'

2. Step 2: Provide a brief explanation of the law's significance

   Explain that the law supports Dalton's atomic theory and the concept of atoms combining in definite ratios.

3. Step 3: Choose an appropriate example

   Select a pair of elements that form at least two different compounds. Common examples include carbon and oxygen (CO, CO₂) or nitrogen and oxygen (N₂O, NO, N₂O₃, NO₂, N₂O₅).

4. Step 4: Analyze the first compound in the example

   For Carbon Monoxide (CO):
   - Identify the mass ratio of elements (C:O = 12:16).
   - Calculate the mass of one element (e.g., oxygen) that combines with a fixed mass (e.g., 1 part) of the other element (carbon). (16/12 = 1.33 parts O per 1 part C).

5. Step 5: Analyze the second compound in the example

   For Carbon Dioxide (CO₂):
   - Identify the mass ratio of elements (C:O = 12:32).
   - Calculate the mass of the same element (oxygen) that combines with the same fixed mass (1 part) of the other element (carbon). (32/12 = 2.66 parts O per 1 part C).

6. Step 6: Demonstrate the simple whole-number ratio

   Compare the calculated masses of oxygen from both compounds (1.33 : 2.66). Show that this ratio simplifies to a simple whole number ratio (1:2), thus confirming the law.