Question

What is the percent by mass of water in MgSO4 • 7H2O? A. 51.1% B. 195% C. 56.0% D. 21.0%

Answer 1

Answer : The correct option is, (A) 51.1%

Explanation :

Mass percent : It is defined as the mass of the given component present in the total mass of the compound.

Formula used :

$\text{Mass} \%H_2O=\frac{\text{Mass of }H_2O}{\text{Mass of }MgSO_4.7H_2O}\times 100$

First we have to calculate the mass of $H_2O$ and $MgSO_4.7H_2O$.

Mass of $H_2O$ = 18 g/mole

Mass of $7H_2O$ = 7 × 18 g/mole = 126 g/mole

Mass of $MgSO_4.7H_2O$ = 246.47 g/mole

Now put all the given values in the above formula, we get the mass percent of $H_2O$ in $MgSO_4.7H_2O$.

$\text{Mass} \%H_2O=\frac{126g/mole}{246.47g/mole}\times 100=51.1$

Therefore, the mass percent of $H_2O$ in $MgSO_4.7H_2O$ is, 51.1%