The number of oxygen atoms in 24.9 g of ${\mathrm{CuSO}}_4\cdot 5{\mathrm{H}}_2\mathrm{O}$ is ( molar mass of $\left.\mathrm{Cu}=63\mathrm{g}{\mathrm{mol}}^{-1}\right)$ 

Molar mass of ${\mathrm{CuSO}}_4\cdot 5{\mathrm{H}}_2\mathrm{O}=(63+32+4\times 16+5\times 18)\mathrm{g}{\mathrm{mol}}^{-1}=249\mathrm{g}{\mathrm{mol}}^{-1}$

Amount of ${\mathrm{CuSO}}_4\cdot 5{\mathrm{H}}_2\mathrm{O}$ in the given mass is $n=\frac{m}{M}=\frac{24.9\mathrm{g}}{249\mathrm{g}{\mathrm{mol}}^{-1}}=0.1\mathrm{mol}$

Number of oxygen atoms $=9\times \left(0.1\mathrm{mol}\right)\left(6.022\times {10}^{23}{\mathrm{mol}}^{-1}\right)=5.42\times {10}^{23}$