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 $60gmo{l}^{-1}$.The osmotic pressure at 313 K, if the solution density at this temperature is 1010 $Kg/m^3$.

• A. $2.56\times {10}^{5}Pa$
• B. $4.45\times {10}^{5}Pa$
• C. $3.76\times {10}^{5}Pa$
• D. None of these

Answer 1

The correct option is A $2.56\times {10}^{5}Pa$

The relative lowering in the vapour pressure is equal to the mole fraction of solute.

$\frac{P^0-P}{P^0}=X_B$

$\frac{2338-2295.8}{2338}=X_B{X}_B=0.018$

Mole fraction of water $=1-0.018=0.982$

We can have a solution of 0.982 moles of water and 0.018 moles of solute. Molar masses of solute and water are 60 g/mol and 18 g/mol respectively.

Mass of solute $60\times 0.018=1.08$g
Mass of water $=18\times 0.982=17.68g$
Total mass of solution $=1.08+17.68=18.76g$

Density of solution is 1010 $kg/m^3=1010g/L$

Volume of solution $=\frac{Mass}{density}=\frac{18.76}{1010}=0.01857L$

The concentration $C=\frac{moles}{Volume}=\frac{0.018}{0.01857}=0.969M$

The osmotic pressure $\Pi =CRT=0.969\times 0.0821\times 313=24.88atm=24.88\times {10}^{5}Pa$