Berapakah titik didih dan titik beku larutan 0,1 mol Ba(NO3)2 dalam 500 gram air. Kb Air = 1,86°C m pangkat -1

n = 0,1 mol
wp = 500 gram
Kb = 1,86
Kd = 0,52
Td = ...........................?
Tb = ...........................?

m = n x 1000/wp
= 0,1 x 1000/500
= 0,1 x 2
= 0,2 molal

ΔTd = m x Kd
= 0,2 x 0,52
= 0,104°C

Td = Td° + ΔTd
= 100 + 0,104
= 100,104°C

ΔTb = m x Kb
= 0,2 x 1,86
= 0,372°C

Tb = Tb° - ΔTb
= 0 - 0,372
= -0,372°C