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  4. 10 g of MnO 2 on reaction with conc HCl liberated 0.1 equivalent of Cl 2( Mn =55). Hence, per cent purity of MnO 2 is

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Q. $10 \,g$ of $MnO _2$ on reaction with conc $HCl$ liberated $0.1$ equivalent of $Cl _2( Mn =55)$. Hence, per cent purity of $MnO _2$ is

Some Basic Concepts of Chemistry

Solution:

$MnO _2+ HCl \rightarrow MnCl _2+ Cl _2+2 H _2 O $
$0.1 $ equivalent $Cl _2=\frac{0.1}{2} mol Cl _2^{-} $
$=0.05\, mol\, MnO _2$
$=4.35\, g \, MnO _2 $ in $ 10 $ g sample
Thus, $43.5 \%$ pure