Fraksi mol suatu larutan methanol (CH3OH) dalam air adalah 0,5. Tentukan kosentrasi methanol dalam larutan ini dinyatakan dalam % massa!

[tex]\displaystyle SIFAT \ KOLIGATIF \ LARUTAN \\ \\ X_t = \frac{nt}{nt+np} \\ 0.5 = \frac{nt}{nt+np} \\ \frac{1}{2} = \frac{nt}{np} \\ \\ nt = 1 \ mol -\ \textgreater \ gr \ CH_3OH \ = 1 \ mol \times 32 \ gram/mol=32 \ gram \\ np = 1 \ mol -\ \textgreater \ gr \ H_2O = 1 \ mol \times 18 \ gram/mol=18 \ gram \\ \% \ CH_3OH = \frac{32}{32+18} \times 100\% = 64\% -\ \textgreater \ Jawab [/tex]

Diketahui :
X CH3OH = 0,5

Mr CH3OH = Ar C + 4 Ar H + Ar O
= 12 + 4(1) + 16
= 32

Mr H2O = 2 Ar H + Ar O
= 2(1) + 16
= 18

Ditanya :
% m/m CH3OH = .... %

X CH3OH = n CH3OH / n total
0,5 = n CH3OH / ( n CH3OH + n air )
0,5 n CH3OH + 0,5 n air = n CH3OH
n air = n CH3OH

Asumsikan :
n air = n CH3OH = y

Maka :
m air = n air × Mr air
= 18y

m CH3OH = n CH3OH × Mr CH3OH
= 32 y

m total = m CH3OH + m air
= 32y + 18y
= 50y

% CH3OH = (m CH3OH / m total) × 100%
= ( 32y / 50y ) × 100%
= 64% m/m