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Industrial Preparation of Ammonia

Equivalent we...

Question

Equivalent weight of $K C l . M g C l_{2} .6 H_{2} O$ is:

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\frac{M}{2}

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\frac{M}{3}

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Solution

The correct option is C $\frac{M}{2}$
$K C l . M g C l_{2} .6 H_{2} O$
Molecular weight=M
$M g$ -oxidation state=+2
Equivalent weight= $\frac{M}{2}$

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