Answer:
104 g of KBr
Step-by-step 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