Em tham khảo nha:
\(\begin{array}{l}
a)\\
{m_{{\rm{dd}}}} = 20 + 180 = 200g\\
{C_\% } = \dfrac{{20}}{{200}} \times 100\% = 10\% \\
b)\\
{n_{N{H_3}}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
{m_{N{H_3}}} = 0,25 \times 17 = 4,25g\\
{m_{{\rm{dd}}}} = 4,25 + 157,5 = 161,75g\\
{C_\% } = \dfrac{{4,25}}{{161,75}} \times 100\% = 2,63\%
\end{array}\)