Em tham khảo nha :
\(\begin{array}{l}
1)\\
C\\
C{\% _{NaCl}} = \dfrac{{30}}{{30 + 170}} \times 100\% = 15\% \\
2)\\
A\\
{m_{BaS{O_4}}} = \dfrac{{100 \times 7}}{{100}} = 7g\\
{m_{{H_2}O}} = 100 - 7 = 93g\\
3)\\
A\\
{n_{CuC{l_2}}} = 0,4 \times 0,2 = 0,08mol\\
{m_{CuC{l_2}}} = 0,08 \times 135 = 10,8g\\
4)\\
B\\
{m_{NaOH}} = \dfrac{{60 \times 30}}{{100}} = 18g\\
C\% = \dfrac{{{m_{NaOH}} + {m_{NaO{H_{ct}}}}}}{{{m_{dd}} + {m_{NaO{H_{ct}}}}}} \times 100\% = 44\% \\
\Rightarrow \dfrac{{{m_{NaO{H_{ct}}}} + 18}}{{{m_{NaO{H_{ct}}}} + 60}} = 0,44\\
\Rightarrow {m_{NaO{H_{ct}}}} = 15g
\end{array}\)