\(\begin{array}{l}
1)\\
nNaOH = m{\rm{dd}} \times C\% = 50 \times 20\% = 10g\\
m{H_2}O = 50 - 10 = 40g\\
2)\\
nNaOH = {C_M} \times V = 0,15 \times 1,5 = 0,225\,mol\\
mNaOH = n \times M = 0,225 \times 40 = 9g\\
3)\\
nNaCl = C\% \times m{\rm{dd}} = 200 \times 30\% = 60g\\
m{\rm{dd}}NaCl = \dfrac{{60}}{{15\% }} = 400g\\
m{H_2}O = 400 - 200 = 200g\\
4)\\
nNaCl = 0,3 \times 2 = 0,6\,mol\\
VNaCl = \dfrac{{0,6}}{1} = 0,6l = 600ml\\
V{H_2}O = 600 - 300 = 300ml
\end{array}\)
