Question

Certain mole of $HCN$ is oxidised completely by $25mL$ of $KMn{O}_4$. The products are $C{O}_2$ and $N{O}_3^{⊝}$ ion. When all $C{O}_2$ is passed through lime water, $1g$ of $CaC{O}_3$ is obtained. The molarity of the $KMn{O}_4$ used is:

• A. $1.44M$
• B. $0.72M$
• C. $0.36M$
• D. none of these

Answer 1

The correct option is D none of these

$HCN+KMn{O}_4\to \underset{\underset{\underset{\left(1g\right)\equiv 0.01mol}{CaC{O}_3}}{\downarrow }}{C{O}_2}+N{O}_3^{⊝}+M{n}^{2+}$
mol of $C{O}_2=0.01$ mol
$C{N}^{⊝}+5H_{2}O\to C{O}_2+N{O}_3^{⊝}+10H^{⨁}+10e^{-}$
$Mn{O}_4^{⊝}+8H^{⨁}+5e^{-}\to M{n}^{2+}+4H_{2}O$
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$C{N}^{⊝}+2Mn{O}_4^{⊝}+6H^{⨁}\to C{O}_2+N{O}_3^{⊝}+2M{n}^{2+}+3H_{2}O$
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$1$ mol $C{O}_2\equiv 2$ mol $Mn{O}_4^{⊝}$
$0.01$ mol $C{O}_2\equiv 0.02$ mol $Mn{O}_4^{⊝}$

$M_{KMn{O}_4}=\frac{\frac{0.02}{25}}{1000}=0.8M$