Question

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

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

Answer 1

Answer: (A)

Relative lowering of $V.P.=\frac{\Delta P}{P^0}=x_{\text{solute }}$

${\left(\frac{\Delta P}{P^0}\right)}_A=\frac{\frac{10}{100}}{\frac{10}{100}+\frac{180}{18}}:{\left(\frac{\Delta P}{P^0}\right)}_B=\frac{\frac{10}{200}}{\frac{10}{200}+\frac{180}{18}}$

${\left(\frac{\Delta P}{P^0}\right)}_C=\frac{\frac{10}{10,000}}{\frac{10}{10,000}+\frac{180}{18}}:{\left(\frac{\Delta P}{P^0}\right)}_A>{\left(\frac{\Delta P}{P^0}\right)}_B>{\left(\frac{\Delta P}{P^0}\right)}_C$