A certain liquid X has a normal boiling point of 129.90°C and a boiling point elevation constant =Kb= 1.67°C·kgmol^−1. Calculate the boiling point of a solution made of 90.g of iron(III) chloride (FeCl3) dissolved in 650.g of X.

Boiling T° of solution = 135.6°C

Explanation:

Formula for elevation boiling point is:

ΔT = Kb . m . i

ΔT =  Boiling point of solution - Boiling point of pure solvent

Kb = Boiling point elevation constant

m = molality →  moles of solute in 1kg of solvent

i = numbers of ions dissolved

FeCl₃ →  Fe³⁺  +  3Cl⁻

In the dissociation of the ionic salt, we determined 4 moles of ions dissolved.

3 for chlorides and 1 for iron. Then i = 4

m → We convert the mass of solute to moles:

90 g . 1mol / 162.2g = 0.555 moles

650 g of solvent = 0.650 kg of solvent

m = 0.555 mol/0.650kg → 0.85

We replace data at formula

Boiling T° of solution - 129.90°C = 1.67°C . kg/mol  .  0.85 mol/kg . 4

Boiling T° of solution = 1.67°C . kg/mol  .  0.85 mol/kg . 4 + 129.90°C

Boiling T° of solution = 135.6°C

the answer is b & c i believe : )

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