Question

An aqueous solution containing 28% by mass of a liquid A (molar mass = 140) has a vapour pressure
of 160 mm at 37°C. The vapour pressure of water at 37°C is 150 mm, then the pressure of pure liquid
A, is -
(A) 180.22 mm (B)300.32 mm (C) 360.15 mm (D) 276.55 mm​

Answer 1

According to Roult's law :

At a given temperature for a solution of volatile liquids, the partial vapour pressure of each component in solution is equal to the product of vapour pressure of pure component and it's mole fraction.

$\sf{P_A\:=\:P^{\circ}\:\times\:x_A}$

$\sf{P_B\:=\:P^{\circ}\:\times\:x_B}$

• A and B are two liquid components
• $\sf{P_A}$ and $\sf{P_B}$ are partial pressures
• $\sf{P^{\circ}_A}$ and $\sf{P^{\circ}_B}$ are pure liquid pressures

According to Dalton, the total pressure is the sum of partial pressures like $\sf{P_A}$, $\sf{P_B}$ ... and so on.

$\sf{P_S\:=\:P _A\:+\:P _B}$

$\sf{P_S\:=\:P^{ \circ} _A\:x_A\:+\:P^{ \circ} _B\:x_B}$

_______________________________

Explanation :

$\sf{P_S\:=\:P^{ \circ} _A\:x_A\:+\:P^{ \circ} _B\:x_B}$ ___ (eq 1)

An aqueous solution contains 28% mass of a liquid A.

Mole fraction of A = $\sf{\frac{28}{140}}$

=> $\sf{0.2}$

Now, liquid B is water.

So, Mass of water = $\sf{100\:-\:28}$

=> $\sf{72}$

Mole fraction of B = $\sf{\frac{72}{100}}$

=> $\sf{4}$

Total number of moles = Mole fraction of A + Mole fraction of B

=> $\sf{0.2\:+\:4}$

=> $\sf{4.2}$

Now,

$\sf{x_A\:=\:\frac{0.2}{4.2}}$

=> $\sf{0.0476}$

$\sf{x_B\:=\:\frac{4.0}{4.2}}$

=> $\sf{0.95}$

Also, given that

$\sf{P_{total}\:=\:160\:mm}$

$\sf{P^{\circ}_B\:=\:150\:mm}$

Substitute the known values in (eq 1)

=> $\sf{160\:=\:P^{\circ}_A\:\times\:0.0476\:+\:0.95\:\times\:150}$

=> $\sf{160\:=\:P^{\circ}_A\:\times\:0.0476\:+\:142.85}$

=> $\sf{P^{\circ}=\:\frac{160\:-\:142.85}{0.0476}}$

$\sf{P^{\circ}=\:360.15\:mm}$

Answer 2

Answer:

Hey mate

Explanation:

Refers with attachment