Question

The vapour pressure of water at $293K$ is $2338Pa$ and the vapour pressure of an aqueous solution is $2295.8Pa$. If solution density is $1010kg/m^3$ at $313K$, calculate the osmotic pressure at $313K$. Molecular weight of solute $=60$

Answer 1

$p^0$ =2338 Pa , $P_s$ = 2295.8 Pa ; T = 293K

$\frac{p^0-P_s}{p^0}=\frac{n\times M}{W}=\frac{n\times M\times 1000}{W\times 1000}=\frac{molality\times M}{1000}$

molality $=\frac{2338-2295.8}{2338}\times \frac{1000}{18}=1.003$

Mass of solution =100 + 1.0030 $\times$ 60 = 1060.1 g = 1.061 kg

Volume of solution = $\frac{1.060}{1010}{m}^3$ = $1.05\times {10}^{-3}{m}^3$

$\therefore molality=\frac{1.003}{1.05\times {10}^{-3}}=955.24mol{m}^{-3}$

Hence , osmotic pressure = $\frac{n}{V}\times R\times T$

$\Rightarrow 955.24\times 8.314\times 313=2.49\times {10}^{6}pa$

Hence osmotic pressure is approximately equal to $2.53\times {10}^6$ .