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%