Question

The molality$\left(m\right)$ of $KMn{O}_4$ in the acidic medium is :
(Density of $KMn{O}_4$ solution $=1.58g$ $m{L}^{-1}$ and $MW$ of $\left(KMn{O}_4\right)=158g$ $mo{l}^{-1}$)

• A. $0.025$
• B. $0.25$
• C. $0.12$
• D. $0.012$

Answer 1

The correct option is D $0.012$

$d_{sol}=M\left(\frac{M{w}_2}{1000}+\frac{1}{m}\right)$
$M=0.02$
$1.58=0.02\left(\frac{158}{1000}+\frac{1}{m}\right)$
$m=0.012$