Question

Rank the following solutions from lowest to highest vapor pressure.

Answer 1

Solution :

When non volatile solute is added to solvent, vapor pressure gets lowered.

Relative lowering in vapor pressure is given :

$\frac{P^0-P}{P^0}$   = $\text{mole fraction}$ of solute

$\frac{P^0-P}{P^0}=x_B$

$P^0$ = vapor pressure of pure solvent

P = vapor pressure of solution

$x_B$ = mole fraction of solute

$x_B=\frac{n_B}{n_A+n_B}$

$n_B$ = $\text{number of moles of solute}$

$n_A$ = $\text{number of moles of solvent}$

Number of moles $=\frac{\text{weight}}{\text{molecular weight}}$

$\frac{P^0-P}{P^0}=\frac{w_B/M_B}{w_A/M_A+w_B/M_B}$

            $\approx \frac{w_B/M_B}{w_A/M_A}$

1. For 10 g of $CH_3COOK$

         $CH_3COOK \rightarrow CH_3COO^- + K^+$

  Ions = 2

It will affect colligative property.

$\frac{P^0-P}{P^0} = \frac{i \times 10/98}{w_A/M_A}$

Relative lowering in vapor pressure will be :

$=\frac{2 \times 10/98}{w_A/M_A}$

$=\frac{0.20}{w_A/M_A}$

2. For 20 g sucrose

Sucrose is non electrolyte, i = 1

$\frac{P^0-P}{P^0} = \frac{ 20/342}{w_A/M_A}$

            $=\frac{0.050}{w_A/M_A}$

3. For 20 g of glucose.

   Glucose is a non electrolyte, i = 1

   $\frac{P^0-P}{P^0} = \frac{20/180}{w_A/M_A}$

               $=\frac{0.11}{w_A/M_A}$

$w_A/M_A$ is same in all three solutions.

Hence, lowering in vapor pressure is maximum in $CH_3COOK$ and minimum is Sucrose.

Vapor pressure from lowest to highest.

10 g of $CH_3COOK$ < 20 g of glucose < 20 g of sucrose