Question

The hydrated salt, $N{a}_{2}S{O}_4.n{H}_{2}O$ undergoes $56$% loss in weight on heating and become anhydrous. The value of $n$ will be:

• A. $5$
• B. $3$
• C. $7$
• D. $10$

Answer 1

The correct option is D $10$

$56$% water is lost and $44$% $N{a}_{2}S{O}_4$ is left.
Molecular weight of $H_{2}O$ is $18$. Molecular weight of $N{a}_{2}S{O}_4$ is $142$.
Ratio of the number of moles is $N{a}_{2}S{O}_4:H_{2}O=\frac{44}{142}:\frac{56}{18}=0.31:3.11=1:10$.
$\therefore$ Value of $n$ in $N{a}_{2}S{O}_4.n{H}_{2}O$ is $10$.