Đáp án:
\(\begin{array}{l}
a)\\
{C_M}A = 0,5M\\
{C_M}B = 0,1M\\
b)\\
{m_{AgCl}} = 5,74g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{n_{HCl\,trong\,{\rm{dd}}A}} = \dfrac{{9,125}}{{36,5}} = 0,25\,mol\\
{n_{HCl\,trong\,{\rm{dd}}\,B}} = \dfrac{{5,475}}{{36,5}} = 0,15\,mol\\
{n_{HCl}} = 0,25 + 0,15 = 0,4\,mol\\
{V_{{\rm{dd}}C}} = \dfrac{{0,4}}{{0,2}} = 2l \Rightarrow {V_1} + {V_2} = 2(1)\\
{C_M}A - {C_M}B = 0,4 \Leftrightarrow \dfrac{{0,25}}{{{V_1}}} - \dfrac{{0,15}}{{{V_2}}} = 0,4\\
\Rightarrow 0,25{V_2} - 0,15{V_1} = 0,4{V_1}{V_2}(2)\\
(1),(2) \Rightarrow 0,25 \times (2 - {V_1}) - 0,15{V_1} = 0,4{V_1}(2 - {V_1})\\
\Rightarrow 0,5 - 0,25{V_1} - 0,15{V_1} = 0,8{V_1} - 0,4{V_1}^2\\
\Rightarrow 0,4{V_1}^2 - 1,2{V_1} + 0,5 = 0\\
\Rightarrow \left[ \begin{array}{l}
{V_1} = 0,5(n)\\
{V_1} = 2,5(l)
\end{array} \right. \Rightarrow {V_2} = 2 - 0,5 = 1,5\,l\\
{C_M}A = \dfrac{{0,25}}{{0,5}} = 0,5M\\
{C_M}B = \dfrac{{0,15}}{{1,5}} = 0,1M\\
b)\\
AgN{O_3} + HCl \to AgCl + HN{O_3}\\
{n_{HCl}} = \dfrac{{0,4}}{{10}} = 0,04\,mol\\
{n_{AgCl}} = {n_{HCl}} = 0,04\,mol\\
{m_{AgCl}} = 0,04 \times 143,5 = 5,74g
\end{array}\)
