Question

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 Pa at 298 K. Calculate:
(i) molar mass of the solute
(ii) vapour pressure of water at 298 K.

Answer 1

(i) Let, the molar mass of the solute be $M\, g mol^{-1}$
$n_1= \frac{90 g}{18 g mol^{-1}} = 5 mol$
Now, the no. of moles of solvent (water),
$n_2= \frac{30 g}{M g mol^{-1}} = \frac{30}{M} mol$
And, the no. of moles of solute,
$P_1 = 2.8kPa$
Applying the relation:
$\frac{(P^o_1-P_1)}{ P^o_1} =\frac{n_2}{(n_1+n_2)}$
$⇒\frac{(P^o_1-2.8)}{ P^o_1} =\frac{\frac{30}{M}}{(\frac{5M+30}{M})}$
$⇒1-\frac{2.8}{ P^o_1} = \frac{30}{(5M+30)}$
$⇒\frac{2.8}{ P^o_1}= \frac{(5M+30-30)}{5M+30}$
$⇒\frac{2.8}{ P^o_1}= \frac{5M}{(5M+30)}$
$⇒ \frac{P^o_1}{2.8}= \frac{(5M+30)}{5M} ...(i)$
After the addition of 18 g of water:
$n_1 = \frac{(90+18g)}{18} =6mol$
$P_1 = 2.9 kPa$
Again, applying the relation:
$\frac{(P^o_1-P_1)}{ P^o_1} =\frac{n_2}{(n_1+n_2)}$
Dividing equation **(i) **by (ii) , we have:
$⇒\frac{2.9}{2.8} = \frac{(\frac{5M+ 30}{5M}) }{(\frac{6M +30}{6M})}$
$⇒\frac{2.9}{2.8} \times\frac{(6M+ 30)}{6} = \frac{(5M +30)}{5}$
$⇒2.9 × 5 × (6M+30) = 2.8 × 6 × (5M+30)$
$⇒87 M +435= 84 M +504$
$⇒ 3M=69$
$⇒M=23u$
Therefore, the molar mass of the solute is $23 g mol^{-1}.$
(ii) Putting the value of $'M'$ in equation (i), we have:
$\frac{P^o_1}{2.8} = \frac{(5\times23 +30)}{(5 \times 23)}$
$⇒\frac{P^o_1}{2.8} =\frac{145}{115}$
$⇒P^o_1=3.53$
Hence, the vapour pressure of water at 298 K is 3.53 kPa.