If 80 g of copper sulphate CuSO₄ . 5H₂O is dissolved in deionised water to make 5L of solution. The concentration of the copper sulphate solution is x $\times$ 10^$-$3 mol L^$-$1. The value of x is _____________.

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

Given, mass of CuSO₄ . 5H₂O = 80 g<br/><br/>The concentration of copper sulphate solution is x $\times$ 10^$-$3 mol/L.<br/><br/>Molarity = ${{Number\,of\,moles\,of\,solute} \over {Volume\,of\,solution(L)}}$ ..... (i)<br/><br/>Molar mass of CuSO₄ . 5H₂O = 63.54 + 32 + 16 $\times$ 4<br/><br/> = 5 $\times$ 18 = 249.54 g/mol<br/><br/>Number of moles of solute = ${{Weight\,of\,solute} \over {Molecular\,mass\,of\,solute}}$<br/><br/>= ${{80g} \over {249.54g/mol}}$ = 0.32 mol<br/><br/>Volume of solution = 5 L<br/><br/>From Eq. (i),<br/><br/>Molarity = ${{0.3205} \over 5}$ = 64.11 $\times$ 10^$-$3 mol/L<br/><br/>$\therefore$ x = 64.11<br/><br/>or x $\approx$ 64<br/><br/>Hence, answer is 64.