For a dilute solution containing 2.5 g of a non-volatile, non- electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2° C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure ( mm Hg ) of the solution is (take Kb=0.76 K kg mol -1 )

Question

For a dilute solution containing 2.5 g of a non-volatile, non- electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2° C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure ( mm Hg ) of the solution is (take Kb=0.76 K kg mol -1 ) (A) 724 (B) 740 (C) 736 (D) 718

Tardigrade Admin · Accepted Answer

When a non-volatile solute is added to pure solvent, there is a decrease in vapour pressure.
So, according to relative lowering in vapour pressure,

\frac{p ^{\circ} - P _{s}}{p ^{\circ}} = x_{2}

where

x_{2}

is the number of moles of solute,

p_{o}

is the vapour pressure of the pure solvent
and

p_{s}

is the vapour pressure of the solution.
Now,

x_{2}

can also represented as

\frac{n _{2}}{( n _{1} + n _{2} )}

.
Thus, we have

\frac{p ^{0} - p _{s}}{p ^{\circ}} = n \frac{n _{2}}{n _{1} + n _{2}}

Neglecting

n_{2}

(as

n_{2} < n 1

) from the denominator, and substituting

p^{\circ} = 1 a t m = 760 mm H g

, we get

\frac{760 - p s}{760} = \frac{2.5/ M _{B}}{100/18} \times \frac{1000}{1000}

= \frac{18}{1000} \times m (1)

According to elevation in boiling point:

Δ T_{b} = K_{b} \times m

\Rightarrow 2 = 0.76 \times m \Rightarrow m = \frac{2}{0.76}

Putting this value in Eq. (1), we get

\frac{760 - p _{s}}{760} = \frac{8}{1000} \times \frac{2}{0.76}

\Rightarrow p_{s} = 760 - \frac{18 \times 2 \times 760}{1000 \times 0.76}

= 760 - 36 = 724 mm H g