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# The hydrated salt,$Na_2SO_4.nH_2O$ undergoes 55.9% loss in weight on heating and becomes anhydrous.The value of n will be

$(a)\;5\qquad(b)\;3\qquad(c)\;7\qquad(d)\;10$

Percent loss of $H_2O$ in one mole of $Na_2SO_4.nH_2O$ is 55.9
$\Rightarrow \large\frac{18n}{142+18n}$$\times 100=55.9$
$\therefore n=10$
Hence (d) is the correct answer.

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