Question

The vapour pressure of water at room temperature is 23.8mm of hg the vapour pressure of an aqueous solution of sucrose with mole fraction 0.1

Answer 1

Answer:- Solution vapor pressure is 21.42 mmHg.

Solution:- From Raoult's law, "Relative lowering of vapor pressure is directly proportional to the mole fraction of the solute."

$\frac{(P^0-P_s)}{P^0}=x$

where, $P^0$ is the vapor pressure of pure solvent, $P_s$ is the vapor pressure of the solution and $x$ is the mole fraction of the solute.

From given data, vapor pressure of the pure solvent is 23.8 mmHg and the mole fraction of the solute is 0.1. We are asked to calculate the vapor pressure of the aqueous solution of glucose. let's plug in the values in the formula:

$\frac{(23.8-P_s)}{23.8}=0.1$

On cross multiply:

$(23.8-P_s)=0.1(23.8)$

$(23.8-P_s)=2.38$

$P_s=23.8-2.38$

$P_s=21.42$

So, the vapor pressure of the aqueous solution of glucose is 21.42 mmHg.

Answer 2

Given that,

$P_{w} = 23.8$

Mole fraction

$x_{2} = 0.1$

We know the relative lowering in vapor pressure

$x_{2} =  \frac{P_{w} - P_{s} }{P_{w} }$

On substituting all values in above formula

$0.1 = \frac{23.8-P_{s} }{23.8}$

$P_{s} = 23.8-2.38 = 21.42 mm Hg$

So, the vapor pressure of solution = 21.42 mm Hg