Question

Suppose you have been given the task of distilling a mixture of hexane + toluene. Pure hexane has a refractive index of 1.375 and pure toluene has a refactive index of 1.497. You collect a distillate sample which has a refractive index of 1.401. Assuming that the refractive index of the hexane + toluene mixture varies linearly with mole fraction, what is the mole fraction of hexane in your sample?

Answer 1

Given:

refractive index of pure hexane, $\mu_{h}$ = 1.375

refractive index of pure hexane, $\mu_{t}$ = 1.497

refractive index of mixture, $\mu$ = 1.401

Formula used:

As refractive index here, behaves like a colligative prop. then using the following formula:

$\mu = n_{h}\mu_{h} + n_{t}\mu_{t}$                         (1)

where,

$\n_{h}$ = hexane  mole fraction

$\n_{t}$ = toulene  mole fraction

$\n_{h}$ + $\n_{t}$ = 1                                      (2)

Solution:

Now, using given formula from eqn (1)

$\mu = n_{h}\mu_{h} + n_{t}\mu_{t}$

$1.401 = 1.375\times n_{h}+ 1.497\times n{t}$              (3)

Multiply eqn (2) by 1.375, we get:

$1.375\times \n_{h}$ + 1.375\times $\n_{t}$ = 1.375         (4)

Now, solving eqn  (3) and (4):

$\n_{t}$ = 0.213

Substituting the value of $\n_{t}$ = 0.213 in eqn (1), we get:

$\n_{h}$ = 1 - 0.213

$\n_{h}$ = 0.787

Therefore, mole fraction of hexane in the sample is $\n_{h}$ = 0.787