Em tham khảo nha :
\(\begin{array}{l}
3)\\
{m_{{\rm{dd}}NaCl}} = \dfrac{{50 \times 100}}{{25}} = 200g\\
{m_{{H_2}O}} = {m_{{\rm{dd}}}} - {m_{NaCl}} = 200 - 50 = 150g\\
4)\\
NaCl\,20\% \\
{m_{NaCl}} = \dfrac{{40 \times 20}}{{100}} = 8g\\
NaCl\,5\% \\
{m_{NaCl}} = \dfrac{{80 \times 5}}{{100}} = 4g\\
C{\% _{NaCl}} = \dfrac{{8 + 4}}{{40 + 80}} \times 100\% = 10\%
\end{array}\)
