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Kohlrausch Law

The equivalen...

Question

The equivalent mass of $H_{3} B O_{3}$ (M = molar mass of $H_{3} B O_{3})$ in its reaction with NaOH to form $N a_{2} B_{4} O_{7}$ is equal to

M

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\frac{M}{2}

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\frac{M}{4}

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\frac{M}{6}

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Solution

The correct option is A $M$
$H_{3} B O_{3}$ is a monobasic acid so equivalent weight $= \frac{M}{1} = M$

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Similar questions

H_{3} B O_{3} (M = m o l a r m a s s o f H_{3} B O_{3})

N a O H

N a_{2} B_{4} O_{7}

H_{2} S O_{4}

H_{2} S O_{4} + N a O H \to N a H S O_{4} + H_{2} O

H_{3} B O_{3}

{N a}_{3} B O_{3}

\frac{M}{3}

H_{3} B O_{3}

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{N a}_{3} B O_{3}

\frac{M}{3}

M

H_{2} S O_{4}

H_{2} S O_{4} + N a O H \to N a H S O_{4} + H_{2} O

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