Em tham khảo nha:
Em xem bổ sung đề câu 5 nhé :
\(\begin{array}{l}
4)\\
{m_{NaOH}} = 500 \times 3\% + 300 \times 10\% = 45g\\
{C_{\% NaOH}} = \dfrac{{45}}{{500 + 300}} \times 100\% = 5,625\% \\
6)\\
{m_{{\rm{dd}}\,NaOH\,35\% }} = 80 \times 1,38 = 110,4g\\
{m_{NaOH}} = 110,4 \times 35\% = 38,64g\\
{m_{{\rm{dd}}NaOH\,2,5\% }} = \dfrac{{38,64 \times 100}}{{2,5}} = 1545,6g\\
{V_{{\rm{dd}}NaOH\,2,5\% }} = \dfrac{{1545,6}}{{1,03}} \approx 1500\,ml
\end{array}\)
