Question

0.1 M solution of glucose was found to be isotonic with a solution of X in 100 g water. Calculate the value to nearest integer of relative lowering in vapour pressure of solution of X in water after multiplying with ${10}^3$.

Answer 1

Since, 0.1 M glucose is isotonic with solution of X in 100 g water.
Thus, conc. of solution of $X=0.1$ mol per litre
$\therefore$ mole of $X=0.1$
Mole of $H_{2}O=1000/18$
$(\because$ Mass of solution $=$ Mass of solvent $=$ Volume of solution for dilute solution)
$\therefore \frac{P^0-P_S}{P^0}=\frac{n}{n+N}=\frac{0.1}{0.1+\left(1000/18\right)}=0.0018$.
The relative lowering in vapour pressure of solution of X in water is 0.0018.