Class 12 Chemistry

# Solution

## In Text Questions 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.

**Answer:** Given,P^{O}_{A} = 450 mm Hg

P^{O}_{B} = 700 mm Hg

P_{Total} = 600 mm Hg

For liquid A and B,

From Rault’s Law,

P_{Total} = P_{A} + P_{B}

= 600 mm Hg = P^{O}_{A}X_{A}+ P^{O}_{B}X_{B}

Since, X_{A} + X_{B} = 1

Hence, X_{B} = 1 – X_{A}

Thus, above equation can be written as

600 mm Hg = P^{O}_{A}X_{A}+ P^{O}_{B}( - X_{A})

Or, 600 = 450 X_{A} + 700(1 – X_{A})

Or, 600 = 450 X_{A} + 700 – 700X_{A}

Or, 600 – 700 = (450 – 700)X_{A}

Or, -100 = -250X_{A}

Or, X_{A} `= (100)/(250)=0.4`

Since X_{B} = 1 – X_{A}

Hence, X_{B} = 1 – 0.4 = 0.6

As P_{A} = P^{O}_{A}X_{A}

So, P_{A} = 450 `xx` 0.4 = 180 mm Hg

Similarly, P_{B} = P^{O}_{A}X_{B}

So, P_{B} = 700 `xx` 0.6 = 420 mm Hg

Now, In vapour phase,

Mole fraction of liquid A

i.e. X_{} `= (P_A)/(P_A+P_B)`

Or, `X_A=(180)/(180+420)=(180)/(600)=0.3`

And mole fraction of liquid B

i.e. `X_B=1-X_A=1-0.3=0.7`

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.

**Answer:**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 `=(W_B)/(M_B)`

Or, n_{urea} `=(50g)/(60g\text(mol)^(-1))=0.83` mol

Now, number of moles of water `=(W_A)/(M_B)`

Or, n_{H2O} `=(850g)/(18g\text(mol)^(-1))=47.2` mol

Now, mole fraction of urea

X_{urea} = n_{urea} ÷ (n_{urea} + n_{H2O})

`=(0.83text(mol))/(0.83text(mol)+47.2text(mol))`

`=(0.83)/(48.03)=0.017`

Now, we know that

X_{urea} = (P^{O}_{A}) ÷ (P^{O}_{A})

Or, `0.017=(23.8-P_A)/(23.8)`

Or, `23.8-P_A=0.017xx23.8=0.40`

Or, `23.8-0.40=P_A`

Or, `P_A=23.40` mm

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.

**Answer:** Given, Elevation of boiling point

ΔT_{b} = T_{b} - T^{O}_{b}

ΔT_{b} = 100°C – 99.63°C

ΔT_{b} = 0.37°C

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,

ΔT_{b} `= (K_b\xx\W_B\xx1000)/(M_B\xx\W_A\gm)`

Or, `0.37=(0.52xx\W_B\xx1000)/(342xx500)`

Or, `W_B=(0.37xx342xx500)/(0.52xx1000)`

`=(126.54)/(0.52xx2)=(63.27)/(0.52)=121.67` g

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.

**Answer:** Given, Lowering of melting point

ΔT_{f} = 1.5°C

K_{f} = 39.5 kg mol^{-1}

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

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

We know that,

ΔT_{f} = `(K_f\xx\W_B\xx1000)/(M_B\xx\W_A\gm)`

Or, `1.5=(3.9xx\W_B\xx1000)/(176xx75g)`

`=(176xx75xx1.5)/(3.9xx1000)`

Or, `W_B=5.077` g

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.

**Answer:** 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, R = 8.314 `xx` 10^{3} Pa K k^{-1} mol^{-1}

Number of moles of the polymer (n) `=1/(185000)` mol

We know that osmotic pressure (Π) `=n/V\RT`

Or, Π `=1/(185000)xx1/(0.45)xx8.314xx10^3xx310`

`=(8314xx310)/(1850xx45)`

`=(2577340)/(83250)=30.95` Pa

Thus, Osmotic pressure = 30.95 Pa

Molality

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