Đáp án:

\( {C_{M{\text{ NaOH}}}}= 3,275M\)

\( C{\% _{NaOH}} = 12\% \)

Giải thích các bước giải:

 Thể tích dung dịch bằng thể tích nước chứ em.

Phản ứng xảy ra:

\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)

\(N{a_2}O + {H_2}O\xrightarrow{{}}2NaOH\)

Ta có:

\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}} \to {{\text{n}}_{Na}} = 2{n_{{H_2}}} = 0,2{\text{ mol}}\)

\( \to {m_{Na}} = 0,2.23 = 4,6{\text{ gam}} \to {{\text{m}}_{N{a_2}O}} = 17 - 4,6 = 12,4{\text{ gam}}\)

\( \to {n_{N{a_2}O}} = \frac{{12,4}}{{23.2 + 16}} = 0,2{\text{ mol}}\)

\( \to {n_{NaOH}} = {n_{Na}} + 2{n_{N{a_2}O}} = 0,2 + 0,2.2 = 0,6{\text{ mol}}\)

\( \to {m_{NaOH}} = 0,6.40 = 24{\text{ gam}}\)

\({V_{dd}} = 183,2ml = 0,1832{\text{ lít}}\)

\( \to {C_{M{\text{ NaOH}}}} = \frac{{0,6}}{{0,1832}} = 3,275M\)

\({m_{{H_2}O}} = 183,2.1 = 183,2{\text{ gam}}\)

BTKL:

\({m_{hh}} + {m_{{H_2}O}} = {m_{dd}} + {m_{{H_2}}}\)

\( \to 17 + 183,2 = {m_{dd}} + 0,1.2\)

\( \to {m_{dd}} = 200{\text{ gam}}\)

\( \to C{\% _{NaOH}} = \frac{{24}}{{200}}.100\%  = 12\% \)