Question

What is the boiling point in ∘C of each of the solutions below? For water, Kb = 0.51(∘C ⋅ kg)/mol.

Part A

A solution of 15.0 g of urea, CH4N2O, in 158.0 g of water at 45.0 ∘C.

Express your answer using five significant figures.

Tb =

nothing

  ∘C  

SubmitRequest Answer

Part B

A solution of 10.0 g of LiCl in 158.0 g of water at 45.0 ∘C, assuming complete dissociation.

Express your answer using five significant figures.

Tb =

nothing

  ∘C  

Answer 1

1. moles of urea = 15/60.055 = 0.25
158 gm water contains 0.25 mole
1000 gm contain (0.25/158) x 1000 = 1.5822
m = 1.5822
delta Tb = Kb m = 0.51 x 1.5822 = 0.8069
So boiling point elevation= 0.8069 °C

Boiling point= 45+ 0.8069 = 45.8069 °C.

2. moles of Licl = 10/ 42.394 = 0.235
158 gm water contains 0.235 mole
1000 gm contain (0.235/158) x 1000 = 1.493
m= 1.493

deltaTb =iKb× m = 2 x0.51 x 1.493 = 1.522
So boiling point elevation= 1.522 °C
(i = 2,because LiCl = Li+ + Cl-)

Boiling point= 45+ 1.522= 46.522 °C