Answer
(a)The percent composition of $CuSO_4$ in $CuSO_4.5H_2O$=63.92%
(b)Percent composition of $CuSO_4$ in 74.4% $CuSO_4.5H_2O$=47.56%
Work Step by Step
(a)The molar mass of $CuSO_4$=molar mass of Cu+ molar mass of S+4$\times$molar mass of O=159.609amu
The molar mass of $CuSO_4.5H_2O$=molar mass of $CuSO_4$+molar mass of $5H_2O$=$159.609+5\times18.015=249.684$
The percent composition of $CuSO_4$ in $CuSO_4.5H_2O$:
percent composition of $CuSO_4$=$\frac{\text{molar mass of $CuSO_4$}}{\text{molar mass of $CuSO_4.5H_2O$}}\times100=\frac{159.609}{249.684}\times100\approx63.92\%$
(b)Percent composition of $CuSO_4$ in 74.4% $CuSO_4.5H_2O$=Percent composition of $CuSO_4$ in $CuSO_4.5H_2O$$\times$ 74.4%$\times100$=$63.92\%\times74.4\%\times100\approx47.56\%$
