Đáp án:
a) 7,97%
b) 1,75 M
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
nNa = \dfrac{m}{M} = \dfrac{{5,75}}{{23}} = 0,25\,mol\\
n{H_2} = \dfrac{{0,25}}{2} = 0,125\,mol\\
m{\rm{dd}} = 5,75 + 120 - 0,125 \times 2 = 125,5g\\
nNaOH = nNa = 0,25\,mol\\
C\% NaOH = \dfrac{{mct}}{{m{\rm{dd}}}} \times 100\% = \dfrac{{0,25 \times 40}}{{125,5}} \times 100\% = 7,97\% \\
b)\\
nMgC{l_2} = \dfrac{m}{M} = \dfrac{{33,25}}{{95}} = 0,35\,mol\\
{C_M}MgC{l_2} = \dfrac{n}{V} = \dfrac{{0,35}}{{0,2}} = 1,75M
\end{array}\)