80 g of copper sulphate $CuSO_4 .5H_2 O$ ....

80 g of copper sulphate $CuSO_4 .5H_2 O$ is dissolved in deionised water to make 5 L of solution. The concentration of the copper sulphate solution is $x \times 10^{-3} mol L^{-1}$. The value of x is _ _ _ _ . (Nearest integer)

[Atomic masses - Cu:63.54 u, S:32 u, O: 16 u, H: 1 u]

Solution

$Molarity = \frac {n}{V} = \frac {w}{M_0 L}$

$\therefore Molarity = \frac {80}{249.54 \times 5 } \approx 64 \times 10^{-3} mol L^{-1}$

$\therefore x = 64$