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,

`\\_(h)` = hexane mole fraction

`\\_(t)` = toulene mole fraction

`\\_(h)` +
`\\_(t)` = 1 (2)

Solution:

Now, using given formula from eqn (1)

`\mu = n_(h)\mu_(h) + n_(t)\mu_(t)`

`1.401 = 1.375* n_(h)+ 1.497* n{t}` (3)

Multiply eqn (2) by 1.375, we get:

`1.375* \\_(h)` + 1.375\times
`\\_(t)` = 1.375 (4)

Now, solving eqn (3) and (4):

`\\_(t)` = 0.213

Substituting the value of
`\\_(t)` = 0.213 in eqn (1), we get:

`\\_(h)` = 1 - 0.213

`\\_(h)` = 0.787

Therefore, mole fraction of hexane in the sample is
`\\_(h)` = 0.787