Đáp án:

 a) $C{\% _B} = \dfrac{{4000{m_1}}}{{23{m_2} + 22{m_1}}}$

b) ${C_M} = \dfrac{{1000{m_1}D}}{{23{m_2} + 22{m_1}}}$

c) $\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{23}}{{228}}$; $D = 0,875(g/ml)$

Giải thích các bước giải:

 a) $2Na + 2{H_2}O \to 2NaOH + {H_2}$

${n_{Na}} = \dfrac{{{m_1}}}{{23}}$

Theo PTHH: ${n_{NaOH}} = {n_{Na}} = \dfrac{{{m_1}}}{{23}};{n_{{H_2}}} = \dfrac{1}{2}{n_{Na}} = \dfrac{{{m_1}}}{{46}}$

Bảo toàn khối lượng: ${m_1} + {m_2} = {m_{ddB}} + {m_{{H_2}}}$

$ \Rightarrow {m_{ddB}} = {m_1} + {m_2} - \dfrac{{{m_1}}}{{46}}.2 = {m_2} + \dfrac{{22}}{{23}}{m_1}$

$ \Rightarrow C{\% _B} = \dfrac{{\dfrac{{{m_1}}}{{23}}.40}}{{{m_2} + \dfrac{{22}}{{23}}{m_1}}}.100\%  = \dfrac{{4000{m_1}}}{{23{m_2} + 22{m_1}}}$

b)

${V_{ddB}} = \dfrac{m}{D} = \dfrac{{{m_2} + \dfrac{{22}}{{23}}{m_1}}}{D} = \dfrac{{23{m_2} + 22{m_1}}}{{23D}}$ (ml)

$ \Rightarrow {C_M} = \dfrac{n}{V} = \dfrac{{\dfrac{{{m_1}}}{{23}}}}{{\dfrac{{23{m_2} + 22{m_1}}}{{23D}}.\dfrac{1}{{1000}}}} = \dfrac{{1000{m_1}D}}{{23{m_2} + 22{m_1}}}$

c) 

$C\%  = 16\%  \Rightarrow \dfrac{{4000{m_1}}}{{23{m_2} + 22{m_1}}} = 16 \Rightarrow \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{23}}{{228}}$

$ \Rightarrow {m_1} = \dfrac{{23}}{{228}}{m_2}$  (1)

${C_M} = 3,5M \Rightarrow \dfrac{{1000{m_1}D}}{{23{m_2} + 22{m_1}}} = 3,5$  (2)

Thay (1) vào (2)

$ \Rightarrow \dfrac{{1000.\dfrac{{23}}{{228}}{m_2}.D}}{{23{m_2} + 22.\dfrac{{23}}{{228}}{m_2}}} = 3,5 \Rightarrow D = 0,875(g/ml)$