Page No. 5

1.1. Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.?

Answer

Mass of solution = Mass of benzene + Mass of carbon tetrachloride

= 22 g + 122 g

= 144 g

Mass percentage of benzene

Mass percentage of carbon tetrachloride

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

Answer

Let the total mass of the solution be 100 g, and the mass of benzene be 30 g.

∴ Mass of carbon tetrachloride = (100 − 30) g

= 70 g

Molar mass of benzene (C₆H₆) = (6×12 + 6×1) g mol⁻¹

= 78 g mol⁻¹

∴ Number of moles of C₆H₆

Molar mass of carbon tetrachloride (CCl₄) = 1×12 + 4×35.5

= 154 g mol⁻¹

∴ Number of moles of CCl₄

= 0.455 mol

Thus, the mole fraction of C₆H₆ is given as:

= 0.458

∴ Mole fraction of CCl₄ = 1 − 0.458 = 0.542

1.3. Calculate the molarity of the following solution: 

(a) 30 g of Co(NO₃)₂⋅6H₂O in 4.3 L of solution.

(b) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL.

Answer

(a) Molar mass of Co(NO₃)₂⋅6H₂O

= 58.7 + 2(14 + 48) + 6×18 g mol⁻¹

= 58.7 + 2(62) + 108 g mol⁻¹

= 58.7 + 124 + 108 g mol⁻¹

= 290.7 g mol⁻¹

∴ Number of moles of Co(NO₃)₂⋅6H₂O

= 0.103 mol

Volume of solution = 4.3 L

Molarity of solution

= 0.024 M

(b) Number of moles present in 1000mL of 0.5 M H₂SO₄ = 0.5 mol

∴ Number of moles present in 30mL of 0.5 M H₂SO₄

= 0.015 mol

Therefore, molarity

= 0.03 M

1.4. Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.

Answer

Moles of urea = 0.25 mol

Mass of solvent (water) = 1 kg = 1000 g

Molar mass of urea (NH₂CONH₂) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16

= 60 g mol⁻¹

∴ Mass of urea in 1000 g of water = 0.25 mol × 60 g mol⁻¹

= 15 g

Total mass of solution = 1000 + 15 g

= 1015 g

= 1.015 kg

Thus, 1.015 kg of solution contain urea = 15 g

∴ 2.5 kg of solution will require urea

= 36.95 g

= 37 g (approximately)

Hence, the mass of urea required is 37 g.

1.5. Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL⁻¹.

Answer

20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.

∴ Mass of solvent (water) = 100 − 20 = 80 g = 0.080 kg

(a) Calculation of morality:

Molar mass of KI = 39 + 127 = 166 g mol⁻¹

Moles of KI

= 0.120 mol

Molality of solution

= 1.5 mol kg⁻¹

(b) Calculation of molarity:

Density of the solution = 1.202 g mL⁻¹

∴ Volume of solution

= 83.2 mL

= 0.0832 L

Molarity of the solution

= 1.44 M

(c) Calculation of the mole fraction of KI:

No. of moles of KI = 0.120 mol

No. of moles of water

= 4.44 mol

Mole fraction of KI

= 0.0263

1.6. 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.

Answer

Solubility of H₂S gas = 0.195 m

∴ Moles of H₂S = 0.195 mol

Mass of water = 1000 g

No. of moles of water

= 55.55 mol

∴ Mole fraction of H₂S gas in the solution (x)

= 0.0035

Pressure at STP = 1 bar

According to Henry’s law,

p_H2S = K_H× x_H2S

= 285.7 bar

1.7. 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.

Answer

Given, K_H = 1.67 × 10⁸ Pa

P_CO2 = 2.5 atm = 2.5×101325 Pa

According to Henry’s law,

P_CO2 = K_H× CO₂

= 1.517 × 10⁻³

We can write,

= 1.517 × 10⁻³…[Since, n_CO2 is negligible as compared to to n_H2O]

For 500 mL of soda water, the volume of water = 500 mL; the mass of water = 500 g,

We can write,

= 500/18 mol of water

= 27.78 mol of water

i.e., n_H2O = 27.78

⇒ n_CO2 = 42.14×10⁻³ mole

Mass of CO₂ = 42.14×10⁻³× 44 g = 1.854 g

1.8. The vapour pressures of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if the total vapour pressure is 600 mm Hg. Also, find the composition of the vapour phase.

Answer

Given,

P_A⁰ = 450 mm Hg

P_B⁰ = 700 mm Hg

P_Total = 600 mm Hg

x_A = ?

Applying Raoult's law,

PA = x_A× P_A⁰

P_B = x_B× P_B⁰ = (1-x_A)P_B⁰

P_Total = P_A + P_B

= x_A× P_A⁰ + (1-x_A)P_B⁰

= P_B⁰ + (P_A⁰-P_B⁰)x_A

Substituting the given values, we get

600 = 700 + (450 − 700)x_A or 250x_A = 100

⇒ x_A = 100/250 = 0.40

Thus, the composition of the liquid mixture will be

x_A (mole fraction of A) = 0.40

x_B (mole fraction of B) = 1 − 0.40 = 0.60

Calculation of composition in the vapour phase:

P_A = x_A×P_A⁰

= 0.40 × 450 mm Hg

= 180 mm Hg

P_B = x_B× P_B⁰

= 0.60 × 700 mm Hg

= 420 mm Hg

Mole fraction of A in the vapour phase

= 0.30

Mole fraction of B in the vapour phase = 1 − 0.30 = 0.70

1.9. Vapour pressure of pure water at 298 K is 23.8 mm Hg. 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.

Answer

Given, P⁰ = 23.8 mm Hg

w₁ = 850 g, M₁ (water) = 18 g mol⁻¹

w₂ = 50 g, M₂ (urea) = 60 g mol⁻¹

Applying Raoult's law,

Thus, the relative lowering of vapour pressure = 0.017

∴ ΔP = 0.017 × 23.8

⇒ P⁰− P_s = 0.017 × 23.8

⇒ P_s = 23.8 − (0.017 × 23.8)

⇒ P_s = 23.4 mm Hg

Thus, the vapour pressure of water in the solution is 23.4 mm Hg.

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

Molal elevation constant for water is 0.52 K kg mol⁻¹.

Answer

Here, elevation of boiling point ΔT_b = (100 + 273) − (99.63 + 273)

= 0.37 K

Mass of water, w₁ = 500 g

Molar mass of sucrose (C₁₂H₂₂O₁₁), M₂ = 11×12 + 22×1 + 11×16

= 342 g mol⁻¹

Molal elevation constant, K_b = 0.52 K kg mol⁻¹

We know that,

= 121.67 g (approximately)

= 122 g

Hence, 122 g of sucrose is to be added.

1.11. 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.

Answer

Lowering in melting point (ΔT_f) = 1.5°C

Mass of solvent (CH₃COOH), w₁ = 75 g

Mass of solute, w₂ = ?

Molar mass of solvent (CH₃COOH), M₁ = 60 g mol⁻¹

Molar mass of solute (C₆H₈O₆), M₂ = 6 × 12 + 8 × 1 + 6 × 16

= 72 + 8 + 96

= 176 g mol⁻¹

 K_f = 3.9 K kg mol⁻¹

Applying the formula,

= 5.08 g (approx)

Hence, 5.08 g of ascorbic acid is needed to be dissolved.

1.12. 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.

Answer

Given,

Number of moles of solute dissolved (n)

V = 450 mL = 0.45 L,

T = 37°C = (37 + 273) K = 310 K

R = 8.314 k Pa L K⁻¹ mol⁻¹ = 8.314 × 10³ Pa L K⁻¹ mol⁻¹

= 30.96 Pa

= 31 Pa (approximately)