Sebanyak 6 gram urea Co(NH2) terlarut dalam 500 Mililiter air.tentukan titik didih larutan (C=12 H=1 O=16 dan kb=0,52°C/m)

∆Tb = m × Kb
∆Tb = mol × 1000/p × Kb
∆Tb = 6/60 × 1000/500 × 0,52
∆Tb = 0,104

maka Tb larutan = 100+0,104 = 100,104°C

wt = 6 gr (Mr = 60)
wp = 500 mL = 500 gr
Kb = 0,52°C
Tb = ........................?

∆Tb = mol × 1000/p × Kb
= 6/60 × 1000/500 × 0,52
= 0,104°C

Tb = Tb° + ΔTb = 100 + 0,104 = 100,104°C