Question

An aqueous solution freezes at $272.4K$, while pure water freezes at $273.0K$, while pure water freezes at $273.0K$. Determine
(a) the molality of the solution
(ii) boiling point of the solution
(iii) lowering of vapour pressure of water at $298K$
(Given, $K_f=1.86Kkgmo{l}^{-1},K_b=0.512Kkgmo{l}^{-1}$ and vapour pressure of water at $298K=22.756$)

Answer 1

(a)$0-T_f=i{K}_{f}m$

$273-272.4=1.86\times m$ or $m=\frac{.6}{1.86}=.32m$

(b) $100-T_b=K_b\times m$

$100-K_b\times m=T_b$ or $T_b=100-.16=99.84$

(c) lowering of vapour pressure $=\frac{P_o-P}{P_o}=\frac{760-22.756}{760}=.97$