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?

A
0.0167
B
0.167
C
1.067
D
10.67

Solution:

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 = $\frac{Moles of solute}{Volume of solution (in L)} = \frac{0 .50}{3} = 0 .167 mol L^{- 1}$

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