The hydrated salt $N a_{2} S O_{4} . n H_{2} O$ undergoes 50.3% loss in weight on heating and become anhydrous. The value of ‘n’ will be__

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answer is 8.

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Detailed Solution

Molar mass of $N a_{2} S O_{4} \cdot n H_{2} O = (142 + 18 n)$
%Loss in mass $= \frac{18 n}{(142 + 18 n)} \times 100$
$50.3 = \frac{18 n \times 100}{142 + 18 n}$
n = 8

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