dik :
m = 1,00 molal
Tfl = -1,91 °C
dit :
% a = ...?
∆Tf = Tfp - Tfl
∆Tf = 0 °C - (-1,91)
∆Tf = 1,91 °C
∆Tf = m × Kf × i
1,91 °C = 1 × 1,86 × i
i = 1,03
HF => H+ + F-
i = 1 + (n - 1) a
1,03 = 1 + (2 - 1) a
a = 0,03
% a = 0,03 × 100 % = 3 %
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Penurunan Titik Beku (∆Tf)
dik :
m = 1,00 molal
Tfl = -1,91 °C
dit :
% a = ...?
∆Tf = Tfp - Tfl
∆Tf = 0 °C - (-1,91)
∆Tf = 1,91 °C
∆Tf = m × Kf × i
1,91 °C = 1 × 1,86 × i
i = 1,03
HF => H+ + F-
i = 1 + (n - 1) a
1,03 = 1 + (2 - 1) a
a = 0,03
% a = 0,03 × 100 % = 3 %