When 105. g of urea CH4N2O are dissolved in 650. g of a certain mystery liquid X, the freezing point of the solution is 2.90°C less than the freezing point of pure X. Calculate the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point.

104 g of KBr

Explanation:

Formula por freezing point depression is:

ΔT = Kf . m . i

where ΔT is the Freezing T° of pure solvent - Freezing T° of solution

In the first case, ΔT = 2.90°C

We need to determine the Kf for X, to apply the formula again and determine the mass of KBr

m = molality (moles of solute in 1kg of solvent)

We convert 650 g to kg → 0.65 kg

We need to determine the moles of urea (solute)

105 g . 1 mol / 60 g = 1.75 mol

m for urea is: 1.75 mol /0.65 kg = 2.69 m

i = number of ions dissolved. Urea is an organic compound, so no ions.

i = 1

2.90°C = Kf . 2.69 m . 1

2.90°C / 2.69 m = 1.08 °C/m

In the next salt, KBr this is a ionic salt that can be dissociated

KBr →  K⁺  +  Br⁻

2 ions are generated, so i = 2

We replace again: 2.90°C = 1.08°C/m . m . 2

2.90°C / (1.08°C/m . 2) = m

molality = 1.34

Moles of solute in 1kg of solvent. Let's find out the moles in the same mass of X, we had before.

1.34 mol/kg . 0.65 kg = 0.873 moles

We convert mass to moles: 0.873 mol . 119 g /mol = 104 g

particle   p is an electron and particle q is a neutron