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Standard XII

Chemistry

Carbon Dioxide

x mL of 0.2...

Question

$x$ mL of $0.24$ M solution of $N a_{2} S O_{3}$ will be oxidized by $180$ mL of $0.32$ M $K M n O_{4}$ in acidic medium. Then, the value of $x$ is

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Solution

The balanced reaction is as follows:
$5 N a_{2} S O_{3} + 3 H_{2} S O_{4} + 2 K M n O_{4} \to 5 N a_{2} S O_{4} + 2 M n S O_{4} + K_{2} S O_{4} + 3 H_{2} O$
$N_{1} V_{1} = N_{2} V_{2}$
$180 \times 0.32 \times 5 = 0.24 \times 2 \times V$
or, $V = 600$ mL.

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x mL of 0.24 M solution of Na2SO3 will be oxidized by 180 mL of 0.32 M KMnO4 in acidic medium. Then, the value of x is

The balanced reaction is as follows:5Na2SO3+3H2SO4+2KMnO4→5Na2SO4+2MnSO4+K2SO4+3H2ON1V1=N2V2180×0.32×5=0.24×2×Vor, V=600 mL.

$x$ mL of $0.24$ M solution of $N a_{2} S O_{3}$ will be oxidized by $180$ mL of $0.32$ M $K M n O_{4}$ in acidic medium. Then, the value of $x$ is

The balanced reaction is as follows:
$5 N a_{2} S O_{3} + 3 H_{2} S O_{4} + 2 K M n O_{4} \to 5 N a_{2} S O_{4} + 2 M n S O_{4} + K_{2} S O_{4} + 3 H_{2} O$
$N_{1} V_{1} = N_{2} V_{2}$
$180 \times 0.32 \times 5 = 0.24 \times 2 \times V$
or, $V = 600$ mL.