Chapter 12, Problem 12.37CP

Interpretation Introduction

(a)

To Determine:

The respective lines of solution in graph.

Calculation:

T_f = K_f I m

As, solution B having the greater value of I = 3 , which indicates that its change in boiling temperature is higher than solution A.
T⁰_f -T_f= T_f^s

With larger value of change in freezing temperature, the temperature of solutionwill be lower of red line.

From graph, blue line having the higher freezing temperature considered as solution A whereas Solution B havethe lower temperature present with red line.

For A, T_f^s = 14 C⁰
For B, T_f^s = 11 C⁰

Interpretation Introduction

(b)

To Determine:

Melting temperature of pure solvent A and B

Calculation:

As, both the solutions having the same concentration
So, considering the concentrations, m₁ = m₂ = 1 m

Freezing point of pure solvent is also considered as the melting point.

T⁰_f =T_f^s + T_f
For A, T_f = K_f I m = 3 * 1 * 1 = 3 C⁰.
T⁰_f = 14 + 3 = 17 C⁰.

For B, T_f = K_f I m = 3 * 3* 1 = 9 C⁰.
T⁰_f = 11 + 9 = 20 C⁰.

Thus, for pure solvent A and B, the melting temperature are 17 C and 20 C

Interpretation Introduction

(c)

To Determine:

Molal concentration of both solution