Hitunglah fraksi mol naftalena C10H8 dan benzena C6H6, apabila larutan itu mengandung 16,8 % berat naftalena.

% naftalena = 16,8%

Mr Naftalena = 10 Ar C + 8 Ar H
= 10(12) + 8(1)
= 128

Mr Benzena = 6 Ar C + 6 Ar H
= 6(12) + 6(1)
= 78

Asumsikan :
m lar = 100 gram
m C10H8 = % C10H8 × m lar
= 16,8% × 100 gr
= 16,8 gram

m C6H6 = m lar - m C10H8
= 100 gr - 16,8 gr
= 83,4 gram

n C10H8 = m / Mr
= 16,8 gr / 128
= 0,13 mol

n C6H6 = m / Mr
= 83,4 gr / 78
= 1,07 mol

n total = n C10H8 + n C6H6
= 0,13 mol + 1,07 mol
= 1,2 mol

X C10H8 = n C10H8 / n tot
= 0,13 mol / 1,2 mol
= 0,1083

X C6H6 = n C6H6 / n tot
= 1,07 / 1,2 mol
= 0,8916