A certain hydrate has the formula MgSO₄.xH₂O. A quantity of 54.2 g of the compound is heated in an oven to drive off the water. If the steam generated exerts a pressure of 24.8 atm in a 2.0 L container at 120^oC, calculate x.

Given wt. =54.2

Molecular wt. =120+18x

Pressure =24.8

Volume =2L

Temperature =393k

Applying ideal gas equation;

pv=nRT

$$\begin{gathered} 24.8\times 2=\frac{\mathrm{molar} \mathrm{mass}}{\mathrm{given} \mathrm{mass}}\times 0.0821\times 393 \\ 24.8\times 2=\frac{54.2\mathrm{x}}{\left(120+18\mathrm{x}\right)}\times 0.0821\times 393 \end{gathered}$$

[Note: Each molecule gives x molecules of water]

After solving we get ; 

x=6.95 $\approx$7