# A solution is 0.5m in $MgSO_4$,0.1M $Alcl_3$ and 0.2M in $(NH_4)_2SO_4.$The total ionic strength is

$(a)\;3.2\qquad(b)\;2.4\qquad(c)\;6.4\qquad(d)\;4.3$

$0.5m$ of $Mg^{2+}$
$0.5m$ of $SO_4^{2-}$
$0.1m$ of $Al^{3+}$
$0.1m$ of $Cl^{-1}$
$0.4m$ of $NH_4^+$
$0.2m$ $SO_4^{2-}$
Ionic strength =$\large\frac{1}{2}$$\sum M_z^2 \Rightarrow \large\frac{1}{2}$$(0.5\times 2^2+0.5\times 2^2+0.1\times 3^2+0.3\times 1^2+0.4\times 1^2+0.2\times 2^2)$
$\Rightarrow 3.2$
Hence (a) is the correct answer.