The hydrated salt, $N a_{2} S O_{4} . n H_{2} O$ undergoes $56$ % loss in weight on heating and become anhydrous. The value of $n$ will be:

5

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3

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7

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10

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Solution

The correct option is D $10$
$56$ % water is lost and $44$ % $N a_{2} S O_{4}$ is left.
Molecular weight of $H_{2} O$ is $18$ . Molecular weight of $N a_{2} S O_{4}$ is $142$ .
Ratio of the number of moles is $N a_{2} S O_{4} : H_{2} O = \frac{44}{142} : \frac{56}{18} = 0.31 : 3.11 = 1 : 10$ .
$∴$ Value of $n$ in $N a_{2} S O_{4} . n H_{2} O$ is $10$ .

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