\(\begin{array}{l}
a)\\
nNaCl = 0,5 \times 2 = 1\,mol\\
mNaCl = 1 \times 58,5 = 58,5g\\
b)\\
mNaCl = 200 \times 10\% = 20g\\
m{H_2}O = 200 - 20 = 180g\\
2)\\
a)\\
nCuS{O_4} = 0,2 \times 0,4 = 0,08\,mol\\
VCuS{O_4} = \dfrac{{0,08}}{2} = 0,04l = 40ml\\
V{H_2}O = 200 - 40 = 160ml\\
b)\\
mCuS{O_4} = 300 \times 10\% = 30g\\
m{\rm{ddCuS}}{{\rm{O}}_4} = \dfrac{{30}}{{50\% }} = 60g\\
m{H_2}O = 300 - 60 = 240g
\end{array}\)
