Question

$13.4$ g of a sample of unstable hydrated salt, $N{a}_{2}S{O}_4\cdot x{H}_{2}O$ was found to contain $6.3$ g of $H_{2}O$. The number of molecules of water of crystalisation is :

• A. $5$
• B. $7$
• C. $2$
• D. $10$

Answer 1

Answer: (B)

$7$

Weight of salt $=13.4$ g.

Weight of $H_{2}O=6.3$ g.

Weight of anhydrous salt $=13.4-6.3=7.1$ g.

Moles of anhydrous salt $=\frac{7.1}{142}=0.05$ mol.

Moles of $H_{2}O=\frac{6.3}{18}=0.35$ mol.

$0.05$ mol of anhydrous salt $=0.35$ mol of $H_{2}O$

$1$ mol of anhydrous salt $=\frac{0.35}{0.05}=7$ mol.

Therefore, the number of molecules of water of crystallisation is $7$.