Respuesta :
Answer:
Mole fraction of ethanol is 0.363.
Mole fraction of methanol is 0.387.
Explanation:
Mole fraction of water =[tex]\chi_1=0.250[/tex]
Mole fraction of ethanol =[tex]\chi_2[/tex]
Mole fraction of methanol = [tex]\chi_3[/tex]
[tex]\chi_1+\chi_2+\chi_3=1[/tex]
[tex]\chi_2+\chi_3=1-\chi_1=0.750[/tex]
[tex]\chi_2+\chi_3=0.750[/tex]
[tex]\chi_1=\frac{n_1}{n_1+n_2+n_3}[/tex]
[tex]\chi_2=\frac{n_2}{n_1+n_2+n_3}[/tex]
[tex]\chi_3=\frac{n_3}{n_1+n_2+n_3}[/tex]
[tex]n_1+n_2+n_3=1[/tex]
Moles of water = [tex]n_1=0.250 mol[/tex]
Moles of ethanol= [tex]n_2[/tex]
Moles of methanol= [tex]n_3[/tex]
[tex]n_2+n_3=0.750 mol[/tex]
Mass of the mixture = M
Mass of water, [tex]m_1[/tex]
[tex]=n_1\times 18.0 g/mol=0.250 mol\times 18.0 g/mol=4.5 g[/tex]
Fraction of water by mass = 0.134
[tex]\frac{n_1\times 18.0 g/mol}{M}=0.134[/tex]
[tex]M=\frac{0.250 mol\times 18.0 g/mol}{0.134}=33.58 g[/tex]
Mass of ethanol = [tex]m_2[/tex]
Mass of methanol = [tex]m_3[/tex]
[tex]m_1+m_2+m_3=M[/tex]
[tex]4.5 g+m_2+m_3=33.58 g[/tex]
[tex]m_2+m_3=29.08 g[/tex]..[1]
[tex]\frac{m_2}{46.0 g/mol}+\frac{m_3}{32.0 g/mol}=0.750 mol[/tex]
[tex]16m_1+23m_3=552[/tex]..[2]
On solving [1] and [2]:
[tex]m_2 = 16.70, m_3= 12.38 g[/tex]
Mole fraction of ethanol =[tex]chi_2[/tex]
[tex]\chi_2=\frac{n_2}{n_1+n_2+n_3}[/tex]
[tex]=\frac{\frac{16.70 g}{46.0 g/mol}}{1 mol}= 0.363[/tex]
Mole fraction of methanol = [tex]chi_3[/tex]
[tex]\chi_3=\frac{n_3}{n_1+n_2+n_3}[/tex]
[tex]=\frac{\frac{12.38 g}{32.0 g/mol}}{1 mol}= 0.387[/tex]
