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

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Solution

The correct option is C 10
100 g of $N a_{2} S O_{4} . n H_{2} O$ contians 56 g of water and 44 g of $N a_{2} S O_{4}$ .

The molar mass of $N a_{2} S O_{4}$ is $23 + 23 + 32 + 4 (16) = 142 g$ .

44 g of $N a_{2} S O_{4}$ corresponds to 56 g of water.

142 g of $N a_{2} S O_{4}$ will correspond to $\frac{142 \times 56}{44} = 180$ g of water.

This corresponds to $\frac{180}{18} = 10$ moles.

Thus, the value of n is 10.

Hence, option D is correct.

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