13.4 gm of a sample of unstable hydrate salt ${\mathrm{Na}}_2{\mathrm{SO}}_4.{\mathrm{nH}}_2\mathrm{O}$ was found to contain 6.3 gm of water.

Determine the number of moles of water of crystallization in one mole of hydrated salt.

$$\begin{gathered} \mathrm{weight} \mathrm{of} \mathrm{hydrated} \mathrm{salt} ,{\mathrm{Na}}_2{\mathrm{SO}}_4.{\mathrm{nH}}_2\mathrm{O} =13.4 \mathrm{gm} \\ \mathrm{weight} \mathrm{of} \mathrm{water}=6.3 \mathrm{gm} \\ \mathrm{weight} \mathrm{of} \mathrm{anhydrous} \mathrm{salt} {\mathrm{Na}}_2{\mathrm{SO}}_4=13.4-6.3=7.1 \mathrm{gr} \\ \mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{anhydrous} \mathrm{salt}=142 \mathrm{gm} \\ \mathrm{When} 7.1 \mathrm{gm} \mathrm{of} \mathrm{anhydrous} \mathrm{salt} \mathrm{interacts} \mathrm{with} 6.3 \mathrm{gm} \mathrm{of} \mathrm{water}, \\ \mathrm{then} 142 \mathrm{gm} \mathrm{of} \mathrm{salt} \mathrm{interacts} \mathrm{with} 126 \mathrm{gm} \mathrm{of} \mathrm{water} \\ \frac{ 142\times 6.3}{7.1}=126 \mathrm{gm} \\ \mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{water}=\frac{ 126}{18}=7 \end{gathered}$$