Question

When $2.46\,g$ of a hydrated salt ($MSO_4.xH_2O$) is completely dehydrated, $1.20\,g$ of anhydrous salt is obtained. If the molecular weight of anhydrous salt is $120\,g\,mol^{-1}$ what is the value of $x$ ?

• A. 2
• B. 4
• C. 5
• D. 6
• E. 7

Answer 1

Answer: (E)

$\underset{2.46\, g}{MSO _{4}} \cdot x H _{2} O \stackrel{\Delta}{\longrightarrow } \underset{ 1.20\, g}{MSO _{4}}+ \underset{ x g}{H _{2} O}$

Molecular weight of $MSO _{4}=120\, g / mol$

Molecular weight of $MSO _{4} \cdot x H _{2} O$

$=120\, g +x \times 18 g$

$(120\, g +x .18\, g )$ of $MSO _{4} \cdot x H _{2} O$ on complete dehydration gives $120\, g$ of $MSO _{4}$.

$1\, g$ gives$=\frac{120}{120+18 \cdot x}$

Then, $2.46\, g$ of $MSO _{4} \cdot x H _{2} O$ gives $\frac{120 \times 2.4\, g }{120+18 x}$ which is equal to $1.20\, g$

$\therefore \frac{120 \times 2.46}{120+18 x} =1.20 $.

$\frac{295.2}{120+18 x} =1.20$

$ 295.2 =1.20 \times 120+1.20 \times 18 x $

$295.2 =144+21.6 x $

$ 295.2-144 =21.6 x$

$ x =\frac{151.2}{21.6}=7 $

$ x =7 $

$\therefore $ The value of $x=7$