Question 1

1 Calculate the mass percentage of benzene $(\mathrm{C_6H_6})$ and carbon tetrachloride $(\mathrm{CCl_4})$ if $22\,\mathrm{g}$ of benzene is dissolved in $122\,\mathrm{g}$ of carbon tetrachloride.

Accepted Answer

Mass percentage of benzene = (Mass of benzene / Total mass of solution) × 100 = (22 / (22 + 122)) × 100 = (22 / 144) × 100 = 15.28%
Mass percentage of carbon tetrachloride = (Mass of CCl4 / Total mass of solution) × 100 = (122 / 144) × 100 = 84.72% — Mass percentage is calculated by dividing the mass of the component by the total mass of the solution and multiplying by 100.

Question 2

2 Calculate the mole fraction of benzene in solution containing $30\%$ by mass in carbon tetrachloride.

Accepted Answer

Assuming 100 g solution:
Mass of benzene = 30 g
Mass of CCl4 = 70 g
Molar mass of benzene (C6H6) = 78 g/mol
Molar mass of CCl4 = 154 g/mol
Moles of benzene = 30 / 78 = 0.3846 mol
Moles of CCl4 = 70 / 154 = 0.4545 mol
Mole fraction of benzene = moles of benzene / (moles of benzene + moles of CCl4) = 0.3846 / (0.3846 + 0.4545) = 0.458 — Mole fraction is calculated by dividing the moles of the component by the total moles of all components in the solution.

Question 3

3 Calculate the molarity of each of the following solutions: (a) $30\,\mathrm{g}$ of $\mathrm{Co(NO_3)_2 \cdot 6H_2O}$ in $4.3\,\mathrm{L}$ of solution (b) $30\,\mathrm{mL}$ of $0.5\,\mathrm{M}\,\mathrm{H_2SO_4}$ diluted to $500\,\mathrm{mL}$.

Accepted Answer

(a) Molar mass of $\mathrm{Co(NO_3)_2 \cdot 6H_2O}$ = Co (58.93) + 2 × (14.01 + 3 × 16.00) + 6 × (2 × 1.008 + 16.00) = 58.93 + 2 × 62.01 + 6 × 18.016 = 58.93 + 124.02 + 108.096 = 291.046 g/mol
Moles of solute = 30 / 291.046 = 0.103 mol
Volume of solution = 4.3 L
Molarity = moles / volume = 0.103 / 4.3 = 0.024 M

(b) Initial moles of $\mathrm{H_2SO_4}$ = Molarity × Volume (in L) = 0.5 × 0.030 = 0.015 mol
Final volume = 500 mL = 0.5 L
Molarity after dilution = moles / final volume = 0.015 / 0.5 = 0.03 M — Molarity is calculated by dividing moles of solute by volume of solution in liters. For dilution, moles remain constant; volume changes.

Question 4

4 Calculate the mass of urea $(\mathrm{NH_2CONH_2})$ required in making $2.5\,\mathrm{kg}$ of 0.25 molal aqueous solution.

Accepted Answer

Molality (m) = moles of solute / mass of solvent (kg)
Given: m = 0.25 mol/kg, mass of solvent = 2.5 kg
Moles of urea = molality × mass of solvent = 0.25 × 2.5 = 0.625 mol
Molar mass of urea = 12.01 × 1 + 1.008 × 4 + 14.01 × 2 + 16.00 × 1 = 60.06 g/mol
Mass of urea = moles × molar mass = 0.625 × 60.06 = 37.54 g — Molality is moles of solute per kg of solvent. Calculate moles from molality and solvent mass, then convert to mass using molar mass.

Question 5

5 Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of $20\%$ (mass/mass) aqueous KI is $1.202\,\mathrm{g}\,\mathrm{mL}^{-1}$.

Accepted Answer

Assuming 100 g solution:
Mass of KI = 20 g
Mass of water = 80 g = 0.080 kg
Density = 1.202 g/mL
Volume of solution = mass / density = 100 / 1.202 = 83.19 mL = 0.08319 L
Molar mass of KI = 39.10 + 126.90 = 166.00 g/mol

(a) Molality = moles of solute / mass of solvent (kg) = (20 / 166) / 0.080 = 0.1205 / 0.080 = 1.506 mol/kg

(b) Molarity = moles of solute / volume of solution (L) = (20 / 166) / 0.08319 = 0.1205 / 0.08319 = 1.449 M

(c) Moles of KI = 20 / 166 = 0.1205 mol
Moles of water = 80 / 18 = 4.444 mol
Mole fraction of KI = moles KI / (moles KI + moles water) = 0.1205 / (0.1205 + 4.444) = 0.0264 — Calculate moles of KI and water, then use definitions of molality, molarity, and mole fraction with given density and mass percentages.

Question 6

H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry's law constant.

Accepted Answer

Given: Solubility of H₂S in water at STP = 0.195 mol/kg (molality)

At STP, pressure P = 1 atm = 1.013 × 10^5 Pa

Henry's law states: C = k_H × P
where C = concentration (mol/L), k_H = Henry's law constant (mol/(L·Pa)), P = partial pressure

Since solubility is given in molality (mol/kg), approximate molality ≈ molarity (mol/L) assuming density ~1 kg/L for dilute solution.

Therefore, C = 0.195 mol/L

Rearranging Henry's law constant:

k_H = C / P = 0.195 mol/L / 1.013 × 10^5 Pa = 1.924 × 10^{-6} mol/(L·Pa)

Hence, Henry's law constant for H₂S in water at STP is approximately 1.92 × 10^{-6} mol/(L·Pa). — Step 1: Identify given data - solubility (0.195 mol/kg), pressure (1 atm = 1.013 × 10^5 Pa).
Step 2: Assume molality ≈ molarity for dilute solution.
Step 3: Use Henry's law C = k_H × P.
Step 4: Rearrange to find k_H = C / P.
Step 5: Substitute values and calculate k_H.
Step 6: Final answer: k_H ≈ 1.92 × 10^{-6} mol/(L·Pa).

Question 7

Henry's law constant for CO₂ in water is 1.67×10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm CO₂ pressure at 298 K.

Accepted Answer

Given:
Henry's law constant, k_H = 1.67 × 10^8 Pa
Pressure, P = 2.5 atm = 2.5 × 1.013 × 10^5 Pa = 2.5325 × 10^5 Pa
Volume of soda water = 500 mL = 0.5 L

Henry's law relates pressure and concentration:

P = k_H × C

Rearranged:

C = P / k_H = (2.5325 × 10^5 Pa) / (1.67 × 10^8 Pa) = 1.517 × 10^{-3} mol/L

Quantity of CO₂ in 0.5 L:

n = C × V = 1.517 × 10^{-3} mol/L × 0.5 L = 7.585 × 10^{-4} mol

Molar mass of CO₂ = 44 g/mol

Mass of CO₂ = n × molar mass = 7.585 × 10^{-4} mol × 44 g/mol = 0.0334 g

Therefore, the quantity of CO₂ dissolved in 500 mL soda water is approximately 0.0334 g. — Step 1: Convert pressure to Pascals.
Step 2: Use Henry's law P = k_H × C to find concentration.
Step 3: Calculate moles of CO₂ in given volume.
Step 4: Convert moles to mass using molar mass.
Step 5: Final answer: 0.0334 g CO₂ in 500 mL soda water.

Question 8

Vapour pressure of pure water at 298 K is 23.8 mmHg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Accepted Answer

Given:
Vapour pressure of pure water, p₁⁰ = 23.8 mmHg
Mass of urea, w₂ = 50 g
Mass of water, w₁ = 850 g
Molar mass of urea, M₂ = 60 g/mol
Molar mass of water, M₁ = 18 g/mol

Step 1: Calculate moles of solute (urea) and solvent (water):

n₂ = w₂ / M₂ = 50 / 60 = 0.8333 mol
n₁ = w₁ / M₁ = 850 / 18 ≈ 47.22 mol

Step 2: Calculate mole fraction of water:

x₁ = n₁ / (n₁ + n₂) = 47.22 / (47.22 + 0.8333) ≈ 47.22 / 48.05 ≈ 0.9827

Step 3: Calculate vapour pressure of water in solution:

p₁ = x₁ × p₁⁰ = 0.9827 × 23.8 = 23.39 mmHg

Step 4: Calculate relative lowering of vapour pressure:

Relative lowering = (p₁⁰ - p₁) / p₁⁰ = (23.8 - 23.39) / 23.8 = 0.41 / 23.8 ≈ 0.0172 or 1.72%

Answer:
Vapour pressure of water in solution = 23.39 mmHg
Relative lowering of vapour pressure = 0.0172 (1.72%) — The vapour pressure lowering is calculated using Raoult's law. First, moles of solute and solvent are calculated. Then mole fraction of solvent is found. Vapour pressure of solvent in solution is product of mole fraction and vapour pressure of pure solvent. Relative lowering is the fractional decrease in vapour pressure.

Question 9

Boiling point of water at 750 mmHg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C?

Accepted Answer

Given:
Boiling point of water at 750 mmHg = 99.63°C
Desired boiling point = 100°C
Mass of water, w₁ = 500 g
Elevation in boiling point, ΔT_b = 100 - 99.63 = 0.37°C

Boiling point elevation constant for water, K_b = 0.52 K kg mol⁻¹ (standard value)

Step 1: Calculate molality (m) using boiling point elevation formula:

ΔT_b = K_b × m
m = ΔT_b / K_b = 0.37 / 0.52 = 0.7115 mol/kg

Step 2: Calculate moles of sucrose required:

Mass of solvent in kg = 500 g = 0.5 kg
Moles of sucrose = m × mass of solvent (kg) = 0.7115 × 0.5 = 0.3557 mol

Step 3: Calculate mass of sucrose:

Molar mass of sucrose = 342 g/mol
Mass of sucrose = moles × molar mass = 0.3557 × 342 = 121.6 g

Answer:
121.6 g of sucrose should be added to 500 g of water to raise boiling point to 100°C. — Boiling point elevation is proportional to molality. Using the elevation and Kb, molality is found. Then moles of solute are calculated from molality and solvent mass. Finally, mass of sucrose is found from moles and molar mass.

Question 10

Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. K_f = 3.9 K kg mol⁻¹.

Accepted Answer

Given:
Mass of solvent (acetic acid), w₁ = 75 g = 0.075 kg
Freezing point depression, ΔT_f = 1.5°C
Freezing point depression constant, K_f = 3.9 K kg mol⁻¹
Molar mass of ascorbic acid, M₂ = 176 g/mol (C6H8O6)

Step 1: Calculate molality (m):

ΔT_f = K_f × m
m = ΔT_f / K_f = 1.5 / 3.9 = 0.3846 mol/kg

Step 2: Calculate moles of solute:

moles = m × mass of solvent (kg) = 0.3846 × 0.075 = 0.02885 mol

Step 3: Calculate mass of ascorbic acid:

mass = moles × molar mass = 0.02885 × 176 = 5.08 g

Answer:
5.08 g of ascorbic acid should be dissolved in 75 g of acetic acid to lower melting point by 1.5°C. — Freezing point depression is proportional to molality. Using ΔT_f and K_f, molality is found. Then moles of solute are calculated from molality and solvent mass. Finally, mass of solute is calculated from moles and molar mass.

Question 11

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.

Accepted Answer

Given:
Mass of polymer, w = 1.0 g
Molar mass, M = 185,000 g/mol
Volume of solution, V = 450 mL = 0.450 L = 0.00045 m³
Temperature, T = 37°C = 310 K

Gas constant, R = 8.314 J mol⁻¹ K⁻¹

Step 1: Calculate moles of polymer:

n = w / M = 1.0 / 185000 = 5.405 × 10⁻⁶ mol

Step 2: Calculate osmotic pressure using formula:

\Pi = (n / V) × R × T

\Pi = (5.405 × 10^{-6} mol) / (0.00045 m^3) × 8.314 × 310

\Pi = (0.01201 mol/m^3) × 8.314 × 310

\Pi = 0.01201 × 2577.34 = 30.95 Pa

Answer:
Osmotic pressure exerted by the solution is approximately 31 Pa. — Osmotic pressure is calculated by \Pi = (n/V)RT. Moles of solute are found from mass and molar mass. Volume is converted to cubic meters. Using R in J/mol K and temperature in Kelvin, osmotic pressure in pascals is calculated.

Question 12

Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Accepted Answer

A solution is a homogeneous mixture of two or more substances. The components of a solution are the solute (substance present in smaller amount) and the solvent (substance present in larger amount). Types of solutions are:
1. Solid solutions (e.g., alloys like brass - copper and zinc)
2. Liquid solutions (e.g., salt dissolved in water)
3. Gaseous solutions (e.g., air - mixture of gases)
Each type is classified based on the physical state of solute and solvent. — A solution is defined as a homogeneous mixture where the solute is uniformly distributed in the solvent. Solutions can be solid, liquid, or gaseous depending on the states of solute and solvent. Examples help illustrate these types.

Question 13

Give an example of a solid solution in which the solute is a gas.

Accepted Answer

An example of a solid solution where the solute is a gas is hydrogen dissolved in palladium metal. Hydrogen atoms occupy interstitial sites in the palladium lattice forming a solid solution. — In this solid solution, the gas (hydrogen) atoms dissolve in the solid metal (palladium) without changing the crystal structure, demonstrating a solid solution with gaseous solute.

Question 14

Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage.

Accepted Answer

(i) Mole fraction (x): The ratio of moles of a component to the total moles of all components in the solution.
$$ x_A = \frac{n_A}{\sum n_i} $$

(ii) Molality (m): Number of moles of solute dissolved in 1 kg of solvent.
$$ m = \frac{\text{moles of solute}}{\text{kg of solvent}} $$

(iii) Molarity (M): Number of moles of solute dissolved in 1 litre of solution.
$$ M = \frac{\text{moles of solute}}{\text{litres of solution}} $$

(iv) Mass percentage (% w/w): Mass of solute divided by total mass of solution multiplied by 100.
$$ \% = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100 $$ — Each term defines a concentration unit used to express the composition of solutions. Mole fraction is dimensionless, molality depends on solvent mass, molarity on solution volume, and mass percentage on mass ratio.

Question 15

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL⁻¹?

Accepted Answer

Given:
Mass % of HNO₃ = 68%
Density (d) = 1.504 g/mL
Calculate molarity (M).

Step 1: Calculate mass of 1 L solution = density × volume = 1.504 g/mL × 1000 mL = 1504 g

Step 2: Mass of HNO₃ in 1 L solution = 68% of 1504 g = 0.68 × 1504 = 1022.72 g

Step 3: Molar mass of HNO₃ = 1 + 14 + (16×3) = 63 g/mol

Step 4: Number of moles of HNO₃ = mass / molar mass = 1022.72 / 63 = 16.23 mol

Step 5: Molarity = moles / volume (L) = 16.23 / 1 = 16.23 M

Therefore, molarity of the solution is approximately 16.2 M. — Molarity is calculated by converting mass percentage and density into moles per litre of solution. The density helps find the total mass of solution per litre, from which the mass of solute is found, then converted to moles.

Objectives

Objectives — Study Notes

Objectives

1.1 Types of Solutions

1.2 Expressing Concentration of Solutions

Practice Questions — Objectives

All 5 Chapters in Chemistry-I