Question

The amount of water present in $1.39$g of green vitriol $(FeS{O}_4.7H_{2}O)$ is:
(Molecular mass of $FeS{O}_4$ is $152$)

• A. $1.15$g
• B. $0.63$g
• C. $0.84$g
• D. $1.02$g

Answer 1

The correct option is B $0.63$g

Weight of $FeS{O}_4.7H_{2}O=152+7\left(18\right)=278$

In $278g⟶126grams$ of $H_{2}O$ is present

In $1.39g⟶?\left(x\right)$

$\Rightarrow x=\frac{1.39\times 126}{278}=0.63g$

$\therefore 0.63grams$ of $H_{2}O$ is present in $1.39g$ of $FeS{O}_4$ .