Question

The ratio of lowering of vapour pressure of $0.1M$ aqueous solutions of $KCl,CuS{O}_{4}andCa{F}_2$ is:
Assume 100% ionisation

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

Answer 1

The correct option is D $2:2:3$

Relative lowering of vapour pressure, :

${\left(\frac{\Delta P}{P_A^{\circ }}\right)}_{KCl}=i_{KCl}\times {\chi }_{KCl}$

${\left(\frac{\Delta P}{P_A^{\circ }}\right)}_{CuS{O}_4}=i_{CuS{O}_4}\times {\chi }_{CuS{O}_4}$

${\left(\frac{\Delta P}{P_A^{\circ }}\right)}_{Ca{F}_2}=i_{Ca{F}_2}\times {\chi }_{Ca{F}_2}$

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


$KCl\to K^{+}+C{l}^{-}{i}_{KCl}=2CuS{O}_4\to C{u}^{2+}+S{O}_4^{2-}{i}_{CuS{O}_4}=2Ca{F}_2\to C{a}^{2+}+2F^{-}{i}_{Ca{F}_2}=3$
$i_{KCl}:i_{CuS{O}_4}:i_{Ca{F}_2}2:2:3$