The number of oxygen atoms in $24.9 \mathrm{~g}$ of $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$ is…
The number of oxygen atoms in $24.9 \mathrm{~g}$ of $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$ is (molar mass of $\mathrm{Cu}=63 \mathrm{~g} \mathrm{~mol}^{-1}$ )
$2.41 \times 10^{24}$
$3.01 \times 10^{24}$
$5.42 \times 10^{23}$
$5.42 \times 10^{24}$
Solution
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 $=9 \times(0.1 \mathrm{~mol})\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}ight)=5.42 \times 10^{23}$
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