A $20.0$ mL solution of $N a_{2} S O_{3}$ required $30$ mL of $0.01$ M $K_{2} C r_{2} O_{7}$ solution for the oxidation to $N a_{2} S O_{4}$ . Hence, molarity of $N a_{2} S O_{3}$ solution is:

0.015

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0.045

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0.030

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0.0225

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Solution

The correct option is B $0.045$ M
The balanced chemical equation is $K_{2} C r_{2} O_{7} + 4 H_{2} S O_{4} + 3 N a_{2} S O_{3} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 N a_{2} S O_{4} + 4 H_{2} O$ .
Hence, $1$ mole of potassium dichromate corresponds to $3$ moles of sodium sulfite.
The number of milimoles of potassium dichromate is $30 \times 0.01 = 0.3$ .
The number of milimiles of sodium sulfite is $3 \times 0.3 = 0.9$ milimoles.
The molarity of sodium sulfite is $\frac{0.9}{20.0} = 0.045$ M.

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