What would be the molality of a solution of cobalt (III) chloride that boiled at 110.1?

The solution is 4.93 molal

Explanation:

We assume that solution is aqueous.

We need to apply the colligative property of elevation of boiling point:

ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of pure solvent

110.1°C - 100°C = 10.1°C

Kb = 0.512 °C/m

m = molality → unknown

i = Van't Hoff factor (numbers of ions dissolved)

We assume 100 % dissociation: CoCl₃ → Co³⁺ + 3Cl⁻       i = 4

We have 1 mol of Co³⁺ + 3 moles of chlorides

We replace data → 10.1°C = 0.512°C/m . m . 4

10.1°C / (0.512 m/°C . 4) = m → 4.93

i think slash and

the ratio of the conjugate base to the acid required to attain the desired ph.