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Equivalent Mass

100 mL of N /...

Question

$100 m L$ of $\frac{N}{5} N a O H$ will neutralize

A

0.062 g

of

H_{3} B O_{3}

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B

0.185 g

of

H_{3} B O_{3}

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C

1.24 g

of

H_{3} B O_{3}

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D

0.031 g

of

H_{3} B O_{3}

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Solution

The correct option is C $1.24 g$ of $H_{3} B O_{3}$
$H_{3} B O_{3}$ is a monobasic acid.
$∴ n_{f_{H_{3} B O_{3}}} = 1$

$Let x g of H_{3} B O_{3} will react with 100 m L of \frac{N}{5} N a O H$ .
$Equivalents of H_{3} B O_{3} = Equivalents of N a O H$
$∴ \frac{x}{62} \times 1 = \frac{1}{5} \times \frac{100}{1000}$
$\Rightarrow x = 1.24 g$

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Equivalent Mass

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