berapakah titik didih larutan yang terbentuk dari 20 gram urea CO(NH2)2 dalam 80 gram air jika diketahui Kb air 0,512 derajat celcius/molal

∆Tb = m × Kb
∆Tb = mol × 1000/p × Kb
∆Tb = 20/60 × 1000/80 × 0,512
∆Tb = 2,133° C

maka titik didih larutan = 100 + 2,133 = 102,133°C

wt = 20 gr (Mr = 60)
wp = 80 gr
Kb = 0,512°C
Tb = ........................?

∆Tb = wt/Mr x 1000/wp x Kb
= 20/60 × 1000/80 × 0,52
= 2,145°C

Tb = Tb° + ΔTb = 100 + 2,145 = 102,145°C