a) \(C\%=\frac{20}{580+20}\cdot100\approx3,33\%\)
b) Ta có: \(\left\{{}\begin{matrix}n_{CuSO_4}=\frac{440}{160}=2,75\left(mol\right)\\V_{CuSO_4}=\frac{440}{1,1}=400\left(ml\right)=0,4l\end{matrix}\right.\)
\(\Rightarrow C_M=\frac{2,75}{0,4}=6,875\left(M\right)\)