Ta có :
\(n_{Na}=\frac{m_1}{23}\left(mol\right)\)
\(PTHH:2Na+2H2O\rightarrow2NaOH+H2\)
________\(\frac{m_1}{23}\)_______________\(\frac{m_1}{23}\)____\(\frac{m_1}{46}\)(mol)
BTKL:\(m_{dd_B}=m_{Na}+m_{H2O}-m_{H2}=m_1+m_2-\frac{2.m_1}{46}=\frac{22}{23}m_1+m_2\left(g\right)\)
a,\(C\%_{dd_B}=\frac{40.\frac{m_1}{23}}{\frac{22m_1}{23}+m_2}.100\%=\frac{4000m_1}{22m_1+23m_2}\left(\%\right)\)
b,\(CM_{dd_B}=\frac{m_1}{23}.\frac{\left(\frac{22}{23}m_1+m_2\right)}{d}=\frac{22m_1^2+m_1.m_2}{529d}\)
c,\(C\%=6\%\Rightarrow\frac{4000m_1}{22m_1+23m_2}=6\Rightarrow4000m_1=132m_1+138m_2\)
\(\Rightarrow3868m_1=138m_2\Rightarrow\frac{m_1}{m_2}=\frac{138}{3868}=\frac{69}{1934}\)
\(CM=3,5M\Rightarrow\frac{22m_1^2+m_1m_2}{529d}=3,5\Rightarrow d=\frac{22m^2_1+m_1m_2}{1851,5}\)(g/mol)
