Question

The ratio of van't Hoff factor for three different aqueous solutions of $KCl,CuS{O}_{4}andCa{F}_2$ respectively is:
Assume 100% ionisation

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

Answer 1

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

For 100% ionisation, van't Hoff factor is equal to the number of particles in which 1 ​molecule of electrolyte dissociates​.
$KCl\to K^{+}+C{l}^{-}$
$i_{KCl}=2$
$CuS{O}_4\to C{u}^{2+}+S{O}_4^{2-}$
$i_{CuS{O}_4}=2Ca{F}_2\to C{a}^{2+}+2F^{-}{i}_{Ca{F}_2}=3$
$i_{KCl}:i_{CuS{O}_4}:i_{Ca{F}_2}2:2:3$