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Equivalent Mass

A 0.1097 g sa...

Question

A $0.1097 g$ sample of $A s_{2} O_{3}$ required $36.10 m L$ of $K M n O_{4}$ solution for its titration. The molarity of $K M n O_{4}$ solution is (Given Molecular mass of $A s_{2} O_{3} = 198$ )

0.02

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0.012

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0.03

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Solution

The correct option is C 0.012
Let, molarity of $K M n O_{4}$ solution be M
$∴$ Eq. of $A s_{2} O_{3}$ = Eq. of $K M n O_{4}$ solution
$\frac{0.1097}{198} \times 4 = \frac{36.10 \times M \times 5}{1000}$ (Equivalent weight $A s_{2} O_{3} = \frac{198}{4}$ )
Molarity = 0.0122 M
Hence, (c) is the correct answer.

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Equivalent Mass

Standard XII Chemistry

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A 0.1097 g sample of As2O3 required 36.10 mL of KMnO4 solution for its titration. The molarity of KMnO4 solution is (Given Molecular mass of As2O3=198)

The correct option is C 0.012Let, molarity of KMnO4 solution be M ∴ Eq. of As2O3 = Eq. of KMnO4solution 0.1097198×4=36.10×M×51000 (Equivalent weight As2O3=1984) Molarity = 0.0122 M Hence, (c) is the correct answer.

A $0.1097 g$ sample of $A s_{2} O_{3}$ required $36.10 m L$ of $K M n O_{4}$ solution for its titration. The molarity of $K M n O_{4}$ solution is (Given Molecular mass of $A s_{2} O_{3} = 198$ )