Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and the other has carbon tetrachloride as the solvent. Determine the molal concentration, m (or b ), and boiling point, T b

Molality of both benzene and carbon tetrachloride is 1.32 m .

boiling point is 83.4396 °C

Explanation:

When a non-volaile solute is added to a solvent, its boiling point increases. The relationship increase in temperature and molality of the solute is given as follows:

ΔT = m × Kb

Boiling point of pure benzene is 80.1 °C.

Molal elevation boiling constant of benzene (k_b) is 2.53 °C/m

Increase in boiling point of benzene = m × 2.53

Therefore, Boiling point of benzene solution = 80.1 + 2.53m

Boiling point of pure carbon tetrachloride is 76.8 °C

Molal elevation boiling constant of carbon tetrachloride (k_b) is 5.03 °C/m

Increase in boiling point of benzene = m × 5.03

Therefore, Boiling point of carbon tetrachloride solution = 76.8 + 5.03m

It is given that boiling point of both the solutions are same therefore,

76.8 + 5.03m = 80.1 + 2.53m

5.03m - 2.53m = 80.1 - 76.8

2.5m = 3.3

m = 1.32  

Molality of both benzene and carbon tetrachloride is 1.32 m

Boiling point of the solutions = 80.1 + 2.53 × 1.32

                                            = 83.4396 °C

the millers tale is the answer.

fe 55.85 g/mol and h2o 18.02 g/mol.