1. Hitunglah titik didih dan titik beku larutan glikol 20% berdasar massa ( mr glikol =62 ; Kb dan Kf air berturut-turut 0,52°c dan 1,86°c) 2. Tentukan tekanan

dimisalkan terdapat 100 gram larutan glikol 20%
massa glikol = 20 gr
maka massa air = 100 -20 = 80 gr
m = [massa glikol/Mr] x [1000/massa air]
= [20/62] x [1000/80]
= 4,032
Tb = 100° + ΔTb
= 100° + [molal x Kb]
= 100° + [4,032 x 0,52]
= 100° + 2,1°
= 102,1°C
Tf = 0° - ΔTf
= 0° - [4,032 x 1,86]
= - 7,5°C
2.
T = 25°C = 25 + 273 = 298 K
R = 0,082
massa urea = 0,6 gr
Mr urea = 60
R = 0,082
volume = 1 liter
i = 1
π = ?
jawab ;
M = (massa urea / Mr) / volume
M = (0,6 / 60) / 1 liter = 0,01 M
π = MRTi
π = 0,01 x 0,082 x 298 x 1
π = 0,24 atm