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Equivalent Mass

Mg can reduce...

Question

$M g$ can reduce $N O_{3}^{-}$ to $N H_{3}$ in basic medium.
$N O_{3}^{-} (a q) + M g (s) + H_{2} O \to M g (O H)_{2} (s) + O H^{-} (a q) + N H_{3} (g)$
A $25.0 m L$ sample of $N O_{3}^{-}$ solution was treated with $M g (s)$ . The $N H_{3} (g)$ was passed into $100 m L$ of $0.15 N H C l$ .
The excess of $H C l$ required $32.10 m L$ of $0.10 N N a O H$ for neutralization. What was the molarity of $N O_{3}^{-}$ ions in the original sample?

0.47125

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0.3025

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3.025

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0.5005

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Solution

The correct option is A $0.47125$
$Milliequivalents of N H_{3} formed = Milliequivalents of H C l used = 100 \times 0.15 - 32.10 \times 0.10 = 11.79$

Here, n-factor of $N H_{3}$ is $1$ (acid - base reaction).

For redox change,

$∴ Milliequivalents of N H_{3} for n-factor 8 = 8 \times 11.79 = 94.32$
$∴ Normality of N O_{3}^{-} = \frac{94.32}{25} = 3.77$
$Molarity of N O_{3}^{-} = \frac{3.77}{8} = 0.47125$

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