Answer :
(a) The mole fraction of aniline in the mixture is 0.345
(b) The relative amounts of the two phases is 0.136
Explanation :
(a) First we have to calculate the moles of aniline and hexane.
[tex]\text{Moles of aniline}=\frac{\text{Mass of aniline}}{\text{Molar mass of aniline}}[/tex]
Molar mass of aniline = 93.13 g/mole
[tex]\text{Moles of aniline}=\frac{42.8g}{93.13g/mole}=0.459mole[/tex]
and,
[tex]\text{Moles of hexane}=\frac{\text{Mass of hexane}}{\text{Molar mass of hexane}}[/tex]
Molar mass of hexane = 86.18 g/mole
[tex]\text{Moles of hexane}=\frac{75.2g}{86.18g/mole}=0.873mole[/tex]
Now we have to calculate the mole fraction of aniline.
[tex]\text{Mole fraction of aniline}=\frac{\text{Moles of aniline}}{\text{Moles of aniline}+\text{Moles of hexane}}[/tex]
[tex]\text{Mole fraction of aniline}=\frac{0.459}{0.459+0.873}=0.345[/tex]
Thus, the mole fraction of aniline in the mixture is 0.345
(b) Applying Lever rule, the ratio of amount of each phase:
[tex]\frac{l_{\alpha}}{l_{\beta}}=\frac{0.345-0.308}{0.618-0.345}=0.136[/tex]
Thus, the relative amounts of the two phases is 0.136