The hydrated salt ${\mathrm{Na}}_2{\mathrm{SO}}_4.{\mathrm{nH}}_2\mathrm{O}$ undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of n will be

Loss in weight is due to ${\mathrm{nH}}_2\mathrm{O}$

$\therefore \left(142+18\mathrm{n}\right)\mathrm{g} \mathrm{of} {\mathrm{Na}}_2{\mathrm{SO}}_4.{\mathrm{nH}}_2\mathrm{O}=18\mathrm{ng} \mathrm{loss} \mathrm{in} \mathrm{weight} \mathrm{of} {\mathrm{H}}_2\mathrm{O}$

$100 \mathrm{g} \mathrm{of} {\mathrm{Na}}_2{\mathrm{SO}}_4.{\mathrm{nH}}_2\mathrm{O}=\frac{18\mathrm{n}\times 100}{142+18\mathrm{n}}$

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

Solve for $\mathrm{n}\Rightarrow \mathrm{n}=9.99\approx 10$