Question

The ratio of osmotic pressure of $0.1M$ aqueous solutions of $K_{2}S{O}_4,NaCl$ i.e. $({\pi }_{K_{2}S{O}_4}:{\pi }_{NaCl})$ is:

Assume 100% ionisation, and both the solutions are at same temperature.

• A. $1:2$
• B. $2:1$
• C. $3:2$
• D. $2:3$

Answer 1

The correct option is C $3:2$

Osmotic pressure:

${\left(\pi \right)}_{K_{2}S{O}_4}=i_{K_{2}S{O}_4}\times C_{K_{2}S{O}_4}\times RT$
${\left(\pi \right)}_{NaCl}=i_{NaCl}\times C_{NaCl}\times RT$

Since concentration is same, ratio of osmotic pressure depends only on van't hoff factor.


$NaCl\to N{a}^{+}+C{l}^{-}{i}_{NaCl}=2K_{2}S{O}_4\to 2K^{+}+S{O}_4^{2-}{i}_{K_{2}S{O}_4}=3$
$i_{K_{2}S{O}_4}:i_{NaCl}=3:2$