Question

if you dissolve libr in chloroform and find that the mole fraction of libr is .091. what is the boiling point of this solution?

Answer 1

The boiling point of the solution with 0.091 mole fraction of LiBr is approximately 62.1 °C.

When LiBr dissolves in chloroform, the boiling point of the solution increases compared to pure chloroform. This phenomenon is known as boiling point elevation. Here's how to calculate the new boiling point:

1. Raoult's Law:

We can use Raoult's law to relate the boiling point elevation to the mole fraction of the solute (LiBr) and the boiling point elevation constant of the solvent (chloroform).

2. Calculation:

Given:

Mole fraction of LiBr (x LiBr) = 0.091

Boiling point of pure chloroform (Tb chloroform) = 61.7 °C

Boiling point elevation constant of chloroform (Kb chloroform) = 0.21 °C/mol fraction

Boiling point of the solution (Tb solution):

Tb solution = Tb chloroform + (Kb chloroform * x LiBr)

= 61.7 °C + (0.21 °C/mol fraction * 0.091)

≈ 62.1 °C

Therefore, the boiling point of the solution with 0.091 mole fraction of LiBr is approximately 62.1 °C.