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Đáp án:

Giải thích các bước giải:

Thí nghiệm 1 :

$n_{Na} =  \dfrac{m_1}{23}(mol)$
$2Na + 2H_2O \to 2NaOH + H_2$

Theo PTHH :

$n_{NaOH} = n_{Na} =  \dfrac{m_1}{23}(mol)$

$\to m_{NaOH} = \dfrac{m_1}{23}.40 = \dfrac{40m_1}{23}(gam)$

$n_{H_2} = 0,5n_{Na} =  \dfrac{m_1}{46}(mol)$

Sau phản ứng :

$m_{dd} = m_{Na} + m_{H_2O} - m_{H_2}$
$= m_1 + x - \dfrac{m_1}{46}.2 = \dfrac{22m_1}{23} + x(gam)$

$\to C\%_{NaOH} = \dfrac{ \dfrac{40m_1}{23} }{ \dfrac{22m_1}{23} + x}.100\% = \dfrac{40m_1}{22m_1+23x}.100\%= y\%$(1)

Thí nghiệm 2 :

$n_{Na_2O} = \dfrac{m_2}{62}(mol)$

$Na_2O + H_2O \to 2NaOH$

$n_{NaOH} = 2n_{Na_2O} = \dfrac{m_2}{31}(mol)$

$\to m_{NaOH} = \dfrac{40m_2}{31}(gam)$

$m_{dd} = m_{Na_2O} + m_{H_2O} = m_2 + x(gam)$

$\to C\%_{NaOH} = \dfrac{ \dfrac{40m_2}{31}}{m_2 + x}.100\% = \dfrac{40m_2}{31m_2 + 31x}.100\%= y\%(2)$
Vì $(1) = (2)$ nên :

$ \dfrac{40m_1}{22m_1 + 23x} = \dfrac{40m_2}{31m_2 + 31x}$
$⇔ \dfrac{m_1}{22m_1 + 23x} = \dfrac{m_2}{31m_2 + 31x}$

$b)$
Với $m_1 = 4,6 ; m_2 = 6,9$ thì :

$ \dfrac{4,6}{22.4,6 + 23x} =  \dfrac{6,9}{31.6,9 + 31x}$
$⇔ x = 17,74$

Suy ra :

$C\%_{NaOH} = \dfrac{40m_2}{31m_2 + 31x}.100\% = y\%$

$⇔ \dfrac{y}{100} = \dfrac{40.6,9}{31.6,9 + 31.17,74}$

$⇔ y = 36,13$