suatu larutan mendidih pada 100,2°c.Tentukan titik beku larutan kb air=0,52°c/m dan kf air=1,86°c/m

Δ Tb = Kb.m
100,2-100 = 0,52.m
0,2=0,52.m
m=0,385 m

Δ Tf = Kf.m
=1,86.0,385
=0,72

titik beku -0,72 celcius

SIFAT KOLIGATIF LARUTAN

ΔTb = Tb - Tb°
ΔTb = 100.2 - 100
ΔTb = 0.2°C

[tex]\displaystyle m = m \\ \frac{\delta \ Tb}{Kb} = \frac{\delta \ Tf}{Kf} \\ \frac{0.2}{0.52} = \frac{Tf\° - Tf}{1.86} \\ Tf = - 0.72 \°C \ -\ \textgreater \ \ Jawab [/tex]