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Molecular Mass

100 mL of NaH...

Question

$100 m L$ of $N a H C_{2} O_{4}$ required $50 m L$ of $0.1 M$ $K M n O_{4}$ solution in acidic medium. Volume of $0.1 M N a O H$ required by $100 m L$ of $N a H C_{2} O_{4}$ is $V m L$ . The value of $\frac{V}{25}$ is

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Solution

Number of meq of $C_{2} O_{4}^{2 -} = M n O_{4}^{-} = 50 \times (0.1 \times 5)$

$\Rightarrow$ $N = \frac{50 \times 0.1 \times 5}{100} = 0.25 N$

$M_{1} (C_{2} O_{4}^{2 -}) = 0.125 (a s C_{2} O_{4}^{2 -})$

$\equiv 0.125 N (a s H^{+})$

$\Rightarrow$ $V_{N a O H} = \frac{100 \times 0.125}{0.1} = 125 m L$
$\frac{125}{5} = 5$

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