Question

A sample of acetone undergoes 40% dimerization. Then mole fraction of the dimer in the final mixture is

Answer 1

say we started with 100 mol of acetone.
among which 40% dimerized.
so what we get now is 60mol monomer and 20mol dimer
i.e. total number of mole=60+20=80mol
i.e. mole fraction of dimer in final solution is = 20/80=0.25

Answer 2

Answer:

The mole fraction of the dimer in the final mixture is 0.25.

Explanation:

Suppose there 100 moles of acetone in mixture

Percentage of acetone  undergone dimerization = 40%

Moles undergone dimerization = 40% of 100 moles :

$\frac{40}{100}\times 100mol= 40 moles$

40 moles are dimerzing, which means that moles of dimer are:

$\frac{40 mol}{2}=20 mol$

Percentage of acetone present as monomer =100% - 40% = 60%

= 60% of 100 moles :

$\frac{60}{100}\times 100mol= 60 moles$

Moles of monomer $n_1= 60 moles$

Moles of dimer = $n_2= 20 mol$

Mole fraction of monomer in mixture:

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{60 mol}{60 mol+20 mol}=0.75$

Mole fraction of dimer in mixture:

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{20 mol}{60 mol+20 mol}=0.25$