Page No 55: - Chapter 2 - Solutions - Exercise Solutions - Ncert Solutions class 12 - Chemistry

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Question 2.9:

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:

It is given that vapour pressure of water,  = 23.8 mm of Hg

Weight of water taken, w₁ = 850 g

Weight of urea taken, w₂ = 50 g

Molecular weight of water, M₁ = 18 g mol⁻¹

Molecular weight of urea, M₂ = 60 g mol⁻¹

Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p₁.

Now, from Raoult’s law, we have:

Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173.

Question 2.10:

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 ΔTb = (100 + 273) − (99.63 + 273)

= 0.37 K

Mass of water, w_l = 500 g

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

= 342 g mol⁻¹

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

We know that:

= 121.67 g (approximately)

Hence, 121.67 g of sucrose is to be added.

Note: There is a slight variation in this Answer and the one given in the NCERT textbook.

Question 2.11:

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. Kf = 3.9 K kg mol⁻¹.

Answer:

Mass of acetic acid, w₁ = 75 g

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

= 176 g mol⁻¹

Lowering of melting point, ΔTf = 1.5 K

We know that:

= 5.08 g (approx)

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

Note: There is a slight variation in this Answer and the one given in the NCERT textbook.

Question 2.12:

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:

It is given that:

Volume of water, V = 450 mL = 0.45 L

Temperature, T = (37 + 273)K = 310 K

Number of moles of the polymer, 

We know that:

Osmotic pressure, 

= 30.98 Pa

= 31 Pa (approximately)