Question

The vapour pressure of water at 293 K is 2338 Pa and the vapour pressure of an aqueous solution is 2295.8 Pa. The molar mass of solute is $60gmo{l}^{-1}$. Osmotic pressure at 313 K, if the solution density at this temperature is $1010kg/m^3$ is :

• A. $2.49\times {10}^{6}Pa$
• B. $5.67\times {10}^{6}Pa$
• C. $4.56\times {10}^{6}Pa$
• D. None of these

Answer 1

The correct option is A $2.49\times {10}^{6}Pa$

$P^0=2338Pa,P_S=2295.8Pa;T=293K$

$\frac{P^o-P_S}{P^o}=\frac{n\times M}{W}=\frac{n\times M\times 1000}{W\times 1000}=\frac{molality\times M}{1000}$

$\therefore Molalit{y}^{\prime }{m}^{\prime }=\frac{2338-2295.8}{2338}\times \frac{1000}{18}=1.003m$

Mass of solution $=100+1.003\times 60=1060.1g=1.061kg$

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

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

$\therefore O.P.=\frac{n}{V}\times S\times T=955.24\times 8.314\times 313=2.49\times {10}^{6}Pa$

Option A is correct.