Question

A set of solutions is prepared using $180g$ of water as a solvent and $10g$ of different non-volatile solutes $A,BandC$. The relative lowering of vapour pressure in the presence of these solutes are in the order [Given, molar mass of $A=100gmo{1}^{-1};B=200gmo{l}^{-1};C=10,000gmo{l}^{-1}]$
(JEE MAIN 2020)

• A. $B>C>A$
• B. $C>B>A$
• C. $A>B>C$
• D. $A>C>B$

Answer 1

The correct option is C $A>B>C$

$\text{Relative lowering in vapour pressure (RLVP)}=\frac{\Delta P}{P_o}=\frac{n}{n+N}$ $n\to \text{moles of solutes}$ $N\to \text{moles of solvent}$ $n_A=\frac{\frac{10}{100}}{\frac{10}{100}+\frac{180}{18}}=\frac{0.1}{10.1}=\frac{1}{101}$ $n_B=\frac{\frac{10}{200}}{\frac{10}{200}+\frac{180}{18}}=\frac{0.05}{10.05}=\frac{1}{201}$ $n_C=\frac{\frac{10}{10000}}{\frac{10}{10000}+\frac{180}{18}}=\frac{{10}^{-3}}{10}$ From the above relation, $RLVP\left(A\right)>RLVP\left(B\right)>RLVP\left(C\right)$ Hence, option (c) is correct.