What are the boiling points and freezing points (in oC)of a solution of 50.3 g of I2in 350 g of chloroform? The kb= 3.63 oC/m, the kf= 4.70 oC/m for CHCl3, and the molar mass of I2is 253.8 g/mol. The normal boiling point for CHCl3is 61.2 oC, and the normal freezing point for CHCl3is -63.5 oC.

F.p   ≈  - 66.16⁰c

B.p  ≈ 59.1⁰c

Explanation:

Normal boiling point = 61.2⁰c

Normal freezing point = -63.5⁰c

Kb = 3.63⁰⁰ c/m

Kf = 4.70⁰ c/m

molar mass = 253.8 g/mol

solvent = 350 g = 0.35 kg

calculate The new freezing point and Boiling point

First we have to find the molality = number of moles / solvent in kg

= 0.198 / 0.35 = 0.5657 M

Hence the new freezing point can be calculated as

Normal freezing temp -  new freezing temp = Kf * molality

= -63.5 - F.p = 4.7 * 0.5657

F.p = -63.5 - 2.6588 ≈  - 66.16⁰c

Boiling point

Normal boiling point - new boiling point = Kb * molality

= 61.2 - B.p = 3.63 * 0.5657

B.p = 61.2 - 2.0535 ≈ 59.1⁰c

output?

explanation: