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Standard XII

Physics

Area and Volume Expansion

A 0.5 g sam...

Question

A $0.5 g$ 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 $K I$ 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 the sample.

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Solution

$30.0 m L 0.1 M N a_{2} S_{2} O_{3} = 30.0 m L 0.1 N N a_{2} S_{2} O_{3}$
$\equiv 30.0 m L 0.1 N I_{2}$
$\equiv 30.0 m L 0.1 N C l_{2}$
$\equiv 30.0 m L 0.1 N M n O_{2}$
Amount of $M n O_{2}$ present $= \frac{N \times E \times V}{1000}$
$= \frac{1}{10} \times \frac{87}{2} \times \frac{30}{1000}$

% $M n O_{2} = \frac{87 \times 30 \times 100}{10 \times 2 \times 1000 \times 0.5} = 26.1$ .

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0.5

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is treated with

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A 0.5 g sample containing MnO2 is treated with HCl liberating Cl2. The Cl2 is passed into a solution of KI and 30.0 cm3 of 0.1 M Na2S2O3 are required to titrate the liberated iodine. Calculate the percentage of MnO2 in the sample.

30.0 mL 0.1 M Na2S2O3=30.0 mL 0.1 N Na2S2O3≡30.0 mL 0.1 N I2≡30.0 mL 0.1 N Cl2≡30.0 mL 0.1 N MnO2Amount of MnO2 present =N×E×V1000=110×872×301000% MnO2=87×30×10010×2×1000×0.5=26.1.

A $0.5 g$ 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 $K I$ 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 the sample.