Question:
If $80 \mathrm{~g}$ of copper sulphate $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$ is dissolved in deionised water to make $5 \mathrm{~L}$ of solution. The concentration of the copper sulphate solution is $\mathrm{x} \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$. The value of $\mathrm{x}$ is
[Atomic masses $\mathrm{Cu}: 63.54 \mathrm{u}, \mathrm{S}: 32 \mathrm{u}, \mathrm{O}: 16 \mathrm{u}, \mathrm{H}: 1 \mathrm{u}$ ]
Solution:
Moles of $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}=\frac{80}{249.54}$
Molarity $=\frac{80}{\frac{249.54}{5}}=64.117 \times 10^{-3}$
Nearest integer, $x=64$
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