Respuesta :
Answer:
The mole fraction composition of the liquid is :
Mole fraction of butane, pentane and hexane are 0.3638,0.3908 and 0.2454 respectively.
Explanation:
Mass of the liquid mixture = 200 g
Percentage of butane = 30%
Mass of butane = [tex]\frac{30}{100}\times 200 g=60 g[/tex]
Moles of butane = [tex]n_1=\frac{60 g}{58 g/mol}=1.0345 mol[/tex]
Percentage of pentane= 40%
Mass of pentane= [tex]\frac{40}{100}\times 200 g=80 g[/tex]
Moles of pentane= [tex]n_2=\frac{80 g}{58 g/mol}=1.1111 mol[/tex]
Percentage of hexane = 100% - 30% - 40% = 30%
Mass of hexane = [tex]\frac{30}{100}\times 200 g=60 g[/tex]
Moles of hexane = [tex]n_2=\frac{60 g}{86 g/mol}=0.6977 mol[/tex]
Mole fraction of butane, pentane and hexane : [tex]\chi_1, \chi_2 \& \chi_3[/tex]
[tex]\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{1.0345 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.3638[/tex]
[tex]\chi_2=\frac{n_2}{n_1+n_2+n_3}=\frac{1.1111 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.3908[/tex]
[tex]\chi_3=\frac{n_1}{n_1+n_2+n_3}=\frac{0.6977 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.2454[/tex]
