Question

Aniline and hexane form partially miscible liquid-liquid mixtures below 69 ºC. When 42.8 g of aniline and 75.2 g of hexane are mixed at 67.5 ºC, two separate liquid phases are formed, with mole fractions of aniline of 0.308 and 0.618. Determine the overall mole fraction of aniline in the mixture, and then use the Lever rule to determine the relative amounts of the two phases.

Answer 1

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.

$\text{Moles of aniline}=\frac{\text{Mass of aniline}}{\text{Molar mass of aniline}}$

Molar mass of aniline = 93.13 g/mole

$\text{Moles of aniline}=\frac{42.8g}{93.13g/mole}=0.459mole$

and,

$\text{Moles of hexane}=\frac{\text{Mass of hexane}}{\text{Molar mass of hexane}}$

Molar mass of hexane = 86.18 g/mole

$\text{Moles of hexane}=\frac{75.2g}{86.18g/mole}=0.873mole$

Now we have to calculate the mole fraction of aniline.

$\text{Mole fraction of aniline}=\frac{\text{Moles of aniline}}{\text{Moles of aniline}+\text{Moles of hexane}}$

$\text{Mole fraction of aniline}=\frac{0.459}{0.459+0.873}=0.345$

Thus, the mole fraction of aniline in the mixture is 0.345

(b) Applying Lever rule, the ratio of amount of each phase:

$\frac{l_{\alpha}}{l_{\beta}}=\frac{0.345-0.308}{0.618-0.345}=0.136$

Thus, the relative amounts of the two phases is 0.136