# Solutions

## Solution of NCERT In Text Questions (Part - 3)

Question - 2.8 The vapour pressure 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 total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Solution:

Given,

For liquid A and B,

From Rault’s Law,

Now, In vapour phase,

Mole fraction of liquid A

And mole fraction of liquid B

Thus, composition in liquid phase, is 0.4 and 0.6

And that in vapour phase, is 0.3 and 0.7

Question - 2.9 Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH_{2 }CONH_{2 }) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Solution:

Weight of Urea (W_{B }) = 50 g

Molar mass (M_{B }) of Urea (NH_{2 }CONH_{ 2}) = 14 + 1 x 2 + 12 +16 +14 + 1 x 2 = 60 g mol ^{– 1}

Weight of water (W_{ A}) = 850 g

Molar mass of water (M_{ A}) = 18 g mol – 1

Vapour pressure of water (P_{ A}^{ o}) = 23.8 mm Hg

Vapour pressure of water in the given solution P_{ A} = ?

Now, number of moles of urea

Now, number of moles of water

Now, Mole fraction of urea

Now, we know that,

Thus, vapour pressure of water in this solution is 23.40 mm of Hg and its relative lowering is 0.017

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.

Solution:

Given,

Elevation of boiling point

Mass of water (W_{ A}) = 500 g

Molar mass (M_{ B}) of sucrose (C_{ 12}H_{22 }O_{11 }) = 12 x 12 + 1 x 22 + 16 x 11 = 342 g mol ^{ – 1}

Molal constant for water (K_{b }) = 0.52 K kg mol ^{ – 1}

Therefore, W_{ B} = ?

We know that,

Thus, 121.67 g of sucrose to be added.

Question – 2.11 - Calculate the mass of ascorbic acid (Vitamin C, C_{6 }H_{8 }O_{ 6}) 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.

Solution:

Given,

Lowering of melting point

Molar mass of ascorbic acid (M_{ B}) = 176 g mol^{ – 1}

Thus, mass of ascorbic (W_{B }) = ?

We know that,

Thus, required mass of ascorbic acid = 5.077 g

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.

Solution:

Given, Mass (W_{ B}) of polymer = 1 g

Molar mass (M_{B }) of polymer = 185000

Volume of water (V) = 450 mL = 0.45 L

Temperature (T) = 37 0C = 37+273 = 310 K

Osmotic pressure = ?

We know,

Number of moles of the polymer (n)

We know that Osmotic pressure

Thus, Osmotic pressure = 30.95 Pa

Solutions - Class 12 chemisty - NCERT In Text Solution (Part 1)

Solutions - Class 12 chemisty - NCERT In Text Solution (Part 2)