Question

Two solutions with the same concentration and the same boiling point, but one has bensene as the solvent and the other has carbon tetrachloride as the solvent. Determine that molal concentration, m (or b), and boiling point, Tb. I have for benzene: normal boiling point; 80.1C, and Kb(C/m); 2.53   For the carbon tetrachloride: normal boiling point; 76.8C and Kb(C/m); 5.03

Answer 1

Expression for elevation of boiling point(ΔT) can be written as follows:

ΔT = m x Kb

Where, m = molality

Kb = ebullioscopic constant

For benzene solution, boiling point can be written as sum of boiling point of pure benzene and elevation in boiling point due to addition of solute.

Boiling point of benzene solution = boiling point of benzene +( ΔT)benzene solution

Boiling point of benzene solution = boiling point of benzene + (m x Kb)

Boiling point of benzene solution = 80.1 + (m x 2.53)--------------------------(1)

Similarly, for carbon tetrachloride solution boiling point can be written as follows:

Boiling point of carbon tetrachloride solution = 76.8 + (m x 5.03)----------(2)

Since, concentrations and boiling points of both solutions are same; equations (1) and (2) can be equated as follows:

80.1 + (m x 2.53) = 76.8 + (m x 5.03)

80.1 - 76.8 = (m x 5.03) - (m x 2.53)

3.3 = m x 2.5

m = 3.3/2.5

m = 1.32 mol/kg

The boiling point can be calculated by substituting value of m in equation (1) as follows:

Boiling point of benzene solution = 80.1 + (1.32 x 2.53)

Boiling point of benzene solution = 80.1 + 3.34

Boiling point of benzene solution = 83.44^oC

Therefore, molality of both solutions is 1.32 mol/kg and boiling point is 83.44^oC