$1.61\mathrm{gm}$ of ${\mathrm{Na}}_2{\mathrm{SO}}_4\cdot 10{\mathrm{H}}_2\mathrm{O}$ contains same number of oxygen atoms as present in:

Number of oxygen atom in given compound,

$\frac{1.61}{322}\times 14=0.07\text{ moles ;}$

(a) $\frac{0.98}{98}\times 4$   (b) $\frac{0,08}{64}\times 3$

(c) $\frac{1.78}{178}\times 7$   (d) $\frac{0.05}{100}\times 3$