Question

An ionic compound $MgC{l}_2$ is $75\%$ ionised in an aqueous solution. The van't Hoff factor $\left(i\right)$ of $MgC{l}_2$ in the solution is:

• A. 1
• B. 1.5
• C. 2
• D. 2.5

Answer 1

The correct option is D 2.5

$MgC{l}_2\left(aq\right)\rightleftharpoons M{g}^{2+}+2C{l}^{-}Initially:C00Eqb:C-C\alpha C\alpha 2C\alpha$

$i=\frac{\text{Toal no. of moles after dissociation}}{\text{No. of moles before dissociation}}$

$i=\frac{C-C\alpha +C\alpha +2C\alpha }{C}i=\frac{C+2C\alpha }{C}=1+2\alpha$
Given that :
$\alpha =0.75$
$i=1+(2\times 0.75)i=2.5$