Question

Match the following aqueous solutions with the appropriate letter from the column on the right. Assume complete dissociation of electrolytes

0.21 m CrSO4

0.16 m CuCl2

0.19 m CuSO4

0.44 m Glucose (non electrolyte)

A. Highest boiling point

B. Second highest boiling point

C. Third highest boiling point

D. Lowest boiling point

Answer 1

ANSWER :

Highest Boiling Point	0.16 m CuCl 2
Second Highest Boiling Point	0.44 m Glucose
Third Highest Boiling Point	0.21 m CrSO4
Lowest Boiling Point	0.19 m CuSO4

We have relation, B.P of solution - B.P of solvent =i K _b m

Where ,i is van't Hoff factor, K _b is boiling point constant of solvent and m is molality of solution.

First calculate i values .

CrSO ₄ Cr ²⁺ + SO ₄^2-

One mole of chromium sulfate produces one mol of Cr ²⁺ and one mol of SO ₄^2- . Hence, i = 2

CuCl ₂ Cu ²⁺ + 2 Cl ^-

One mole of copper chloride produces one mol of Cu ²⁺ and two mol of Cl ^- . Hence, i = 3

CuSO ₄ Cu ²⁺ + SO ₄^2-

One mole of copper sulfate produces one mol of Cu ²⁺ and one mol of SO ₄^2- . Hence, i = 2

Glucose is non electrolyte , hence i = 1 .

Consider relation, B.P of solution - B.P of solvent =i K _b m

B.P of solution -100 ⁰ C = 2 0.512 ⁰ C/m 0.21 m

B.P of solution -100 ⁰ C = 0.215  ⁰ C

B.P of solution = 100 ⁰ C + 0.215  ⁰ C

B.P of solution = 100.215 ⁰ C

B.P of Copper chloride solution

Consider relation, B.P of solution - B.P of solvent =i K _b m

B.P of solution -100 ⁰ C = 3 0.512 ⁰ C/m 0.16 m

B.P of solution -100 ⁰ C = 0.246 ⁰ C

B.P of solution = 100 ⁰ C + 0.246 ⁰ C

B.P of solution = 100.246 ⁰ C

B.P of Copper Sulfate solution

Consider relation, B.P of solution - B.P of solvent =i K _b m

B.P of solution -100 ⁰ C =2 0.512 ⁰ C/m 0.19 m

B.P of solution -100 ⁰ C = 0.194 ⁰ C

B.P of solution = 100 ⁰ C + 0.194 ⁰ C

B.P of solution = 100.194 ⁰ C

B.P of Glucose solution

Consider relation, B.P of solution - B.P of solvent =i K _b m

B.P of solution -100 ⁰ C =1 0.512 ⁰ C/m 0.44 m

B.P of solution -100 ⁰ C = 0.225 ⁰ C

B.P of solution = 100 ⁰ C + 0.225 ⁰ C

B.P of solution = 100.225 ⁰ C