0.5 g sample containing $M n O_{2}$ is treated with $H C l,$ liberating $C l_{2}$ . The $C l_{2}$ is passed into a solution of KI and $30.0 c m^{3}$ of $0.1 M N a_{2} S_{2} O_{3}$ are required to titrate the liberated iodine. Calculate the percentage of $M n O_{2}$ in sample. (At. Wt. of $M n = 55$ )

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80.2

26.1

answer is B.

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Detailed Solution

$M n O_{2} \overset{H C l}{\to} C l_{2} \overset{K I}{\to} I_{2} \overset{N a_{2} S_{2} O_{3}}{\to} N a I + N a_{2} S_{4} O_{6}$
Redox changes are: $2 e + I_{2}^{0} \overset{}{\to} 2 I^{-}$
   $2 {(S^{2 +})}_{2} \overset{}{\to} {(S^{5 / 2 +})}_{4} + 2 e$
   $2 e + M n^{4 +} \overset{}{\to} M n^{2 +}$
The reactions suggest that,
Meq. Of    $M n O_{2} = M e q . o f C l_{2}$ formed
$= M e q . o f I_{2}$ liberated
   $= M e q .$ of $N a_{2} S O_{2} O_{3}$ used
$∴ \frac{w}{M / 2} \times 1000 = 0.1 \times 1 \times 30$
$[∵ N_{N a_{2} S_{2} O_{3}} = M_{N a_{2} S_{2} O_{3}} \sin c e v a l e n c y f a c t o r = 1, s e e r e d o x c h a n g e s f o r N a_{2} S_{2} O_{3}]$
Or $w = \frac{0.1 \times 1 \times 30 \times M}{2000} = \frac{0.1 \times 1 \times 30 \times 87}{2000} (∵ M_{M n O_{2}} = 87)$
   $w_{M n O_{2}} = 0.1305$
$∴$ Purity of $M n O_{2} = \frac{0.1305}{0.5} \times 100 = 26.1 %$