Question

When $2.46g$ of a hydrated salt $\left(MS{O}_{4}x{H}_{2}O\right)$ is completely dehydrated, $1.20g$ of anhydrous salt is obtained. If the molecular weight of anhydrous salt is $120g$ ${mol}^{-1}$, what is the value of $x$?

• A. $2$
• B. $4$
• C. $5$
• D. $6$
• E. $7$

Answer 1

Answer: (E)

$7$

$\begin{array}{c}MS{O}_4\\ 2.46g\end{array}\begin{array}{c}\cdot x{H}_{2}O\\ \end{array}\begin{array}{c}\underset{}{\overset{\Delta }{\to }}\\ \end{array}\begin{array}{c}MS{O}_4\\ 120g\end{array}\begin{array}{c}+\\ \end{array}\begin{array}{c}{H}_{2}O\\ xg\end{array}$
Molecular weight of $MS{O}_4=120g/mol$
Molecular weight of $MS{O}_4\cdot x{H}_{2}O=120g+x\times 18g$
$(120g+x\cdot 18g)$ of $MS{O}_4\cdot x{H}_{2}O$ on complete dehydration gives $120g$ of $MS{O}_4$
$1g$ gives $=\frac{120}{120+18\cdot x}$
Then, $2.46g$ of $MS{O}_4\cdot x{H}_{2}O$ gives $\frac{120\times 2.4g}{120+18x}$ which is equal to $1.20g$
$\therefore \frac{120\times 2.46}{120+18x}=1.20$
$\frac{295.2}{120+18x}=1.20$
$295.2=1.20\times 120+1.20\times 18x$
$295.2=144+21.6x$
$295.2-144=21.6x$
$x=\frac{151.2}{21.6}=7$
$x=7$
$\therefore$ The value of $x=7$