What is the concentration of copper sulphate (in mol L^-1), if 80 g of it is dissolved in enough water to make a final volume of 3 L?

Molar mass of CuSO₄ = 63.5 + 32 + 64 = 159.5
Moles of CuSO₄ $=\frac{80}{159.5}=0.50$
Volume of solution = 3 L
Molarity = $\frac{ \mathrm{Moles} \mathrm{of} \mathrm{solute} }{ \mathrm{Volume} \mathrm{of} \mathrm{solution} \left(\mathrm{in} \mathrm{L}\right)}=\frac{0.50}{3}=0.167 \mathrm{mol} {\mathrm{L}}^{-1}$