Hitunglah fraksi mol 24% CH3COOH (C=12,O=16,H=1) dalam air!

Asumsikan :
m CH3COOH = y

% CH3COOH = 24%
(m CH3COOH / (m CH3COOH + m air) ) × 100% = 24%
y / ( y + m air ) = 0,24
y = 0,24 y + 0,24 m air
0,76 y = 0,24 m air
m air = 3,1667 y

Mr CH3COOH = 2 Ar C + 4 Ar H + 2 Ar O
= 2(12) + 4(1) + 2(16)
= 60

n CH3COOH = m / Mr
= y gram / 60
= 0,01667 y mol

Mr Air = 2 Ar H + Ar O
= 2(1) + 16
= 18

n Air = m / Mr
= 3,1667 y / 18
= 0,176 y mol

n total = n air + n CH3COOH
= 0,176y mol + 0,01667y mol
= 0,1926 y mol

X CH3COOH = n CH3COOH / n total
= 0,01667 y mol / 0,1926 y mol
= 0,0865