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Equivalent Mass

A 10 g sample...

Question

A $10 g$ sample of $C u S$ and $C u_{2} S$ was treated with $100 m L$ of $1.25 M K_{2} C r_{2} O_{7}$ to produce $C r^{3 +}, C u^{2 +}$ and $S O_{2}$ . The excess oxidant was reacted with $50 m L$ of $F e^{2 +}$ solution. $25 m L$ of the same $F e^{2 +}$ solution required $0.875 M K M n O_{4}$ under acidic condition, the volume of $K M n O_{4}$ used was $20 m L$ . Find the percentage of $C u_{2} S$ in the sample.
Given: molar mass of $C u$ is $63.5 g m o l^{- 1}$

A

28.7 %

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B

71.3 %

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C

57.4 %

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D

42.6 %

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Solution

The correct option is D $42.6 %$
Redox changes are,
$+ 6 C r_{2} O_{7}^{2 -} \to + 3 C r^{3 +}$ (n-factor = 6)
$+ 7 M n O_{4}^{-} \to + 2 M n^{2 +}$ (n-factor = 5)
$+ 2 C u - 2 S \to + 2 C u^{2 +} + + 4 S O_{2}$ (n-factor = 6)
$+ 1 C u_{2} - 2 S \to + 2 C u^{2 +} + + 4 S O_{2}$ (n-factor = 8)

$Equivalents of dichromate present initially = \frac{1.25 \times 6 \times 100}{1000} = 0.75$
$Equivalents of F e^{2 +} in 25 m L = \frac{0.875 \times 5 \times 20}{1000} = 0.0875$
$Equivalents of F e^{2 +} in 50 m L = 0.0875 \times 2 = 0.175$
$∴ Equivalents of excess dichromate = 0.175$
$Equivalents of dichormate consumed by C u S and C u_{2} S = 0.75 - 0.175 = 0.575$
$∴ Equivalents of C u S + Equivalents of C u_{2} S = 0.575$

If $x g$ is the mass of $C u S,$ the mass of $C u_{2} S$ is $(10 - x) g$ .

$\frac{x}{63.5 + 32} \times 6 + \frac{(10 - x)}{2 (63.5) + 32} \times 8 = 0.575$
$∴ x = 5.74 g$
$% C u S = \frac{5.74}{10} \times 100 = 57.4 %$
$% C u_{2} S = \frac{(10 - 5.74)}{10} \times 100 = 42.6 %$

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Equivalent Mass

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