Đáp án:

Bạn tham khảo lời giải ở dưới nhé !

Giải thích các bước giải:

\(Na + {H_2}O \to NaOH + \dfrac{1}{2}{H_2}\)

\(\begin{array}{l}
1.\\
{n_{Na}} = \dfrac{{{m_1}}}{{23}}(mol)\\
 \to {n_{NaOH}} = {n_{Na}} = \dfrac{{{m_1}}}{{23}}(mol)\\
 \to {m_{NaOH}} = \dfrac{{{m_1}}}{{23}} \times 40(g)
\end{array}\)

\(\begin{array}{l}
{n_{{H_2}}} = \dfrac{1}{2}{n_{Na}} = \dfrac{{{m_1}}}{{46}}(mol) \to {m_{{H_2}}} = \dfrac{{2{m_1}}}{{46}}(g)\\
 \to {m_{{\rm{dd}}b}} = {m_{Na}} + {m_{{H_2}O}} - {m_{{H_2}}} = {m_1} + {m_2} - \dfrac{{2{m_1}}}{{46}}\\
 \to C{\% _b} = \dfrac{{\dfrac{{{m_1}}}{{23}} \times 40}}{{{m_1} + {m_2} - \dfrac{{2{m_1}}}{{46}}}}(\% )
\end{array}\)

\(\begin{array}{l}
2.\\
{V_{{\rm{ddb}}}} = \dfrac{{{m_{{\rm{dd}}}}}}{D} = \dfrac{{{m_1} + {m_2} - \dfrac{{2{m_1}}}{{46}}}}{{1,2}}(l)\\
 \to C{M_b} = \dfrac{n}{V} = \dfrac{{\dfrac{{{m_1}}}{{23}}}}{{\dfrac{{{m_1} + {m_2} - \dfrac{{2{m_1}}}{{46}}}}{{1,2}}}}(M)
\end{array}\)