Question
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
Answer :
0.167
Solution :
$$\eqalign{
& {\text{Molar mass of}} \cr
& CuS{O_4} = 63.5 + 32 + 64 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 159.5 \cr
& {\text{Moles of}}\,\,CuS{O_4} = \frac{{80}}{{159.5}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.50 \cr
& {\text{Volume of solution}} = 3\,L \cr
& {\text{Molarity}} = \frac{{{\text{Moles of solute}}}}{{{\text{VoIume of solution}}\,\,{\text{in}}\,\,L}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.50}}{3} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.167\,mol\,{L^{ - 1}} \cr} $$