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The hydrated salt ${Na}_{2} {SO}_{4} . {nH}_{2} O$ undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of n will be

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Correct option is D

Loss in weight is due to ${nH}_{2} O$

$∴ (142 + 18 n) g of {Na}_{2} {SO}_{4} . {nH}_{2} O = 18 ng loss in weight of H_{2} O$

$100 g of {Na}_{2} {SO}_{4} . {nH}_{2} O = \frac{18 n \times 100}{142 + 18 n}$

$∴ \frac{18 n \times 100}{142 + 18 n} = 55.9$

Solve for $n \Rightarrow n = 9.99 \approx 10$

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The hydrated salt Na2SO4.nH2O undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of n will be

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The hydrated salt ${Na}_{2} {SO}_{4} . {nH}_{2} O$ undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of n will be