Answer:
We need 135.5 grams of potassium bromide
Step-by-step explanation:
Step 1: Data given
Mass of urea = 130 grams
Mass of liquid = 950 grams = 0.950 kg
The freezing point of the solution is 2.90 C less than the freezing point of pure X
The van't Hoff factor = 1.9 for potassium bromide in X
Step 2: Calculate moles urea
Moles urea = mass urea / molar mass urea
Moles urea = 130 grams / 60.06 g/mol
Moles urea = 2.16 moles
Step 3: Calculate molality
Molality = moles urea / mass liquid X
Molality = 2.16 moles / 0.950 kg
Molality = 2.27 molal
Step 4: Calculate Kf
ΔT = i*Kf*m
⇒with ΔT = the freezing point depression = 2.90 °C
⇒with i = the van't Hoff factor of urea = 1
⇒with Kf = the freezing point depression constant = TO BE DETERMINED
⇒with m = the molality = 2.27 molal
2.90 °C = 1 * Kf * 2.27 molal
Kf = 2.90 / 2.27
Kf = 1.28 °C/m
Step 5: Calculate molality
ΔT = i*Kf*m
⇒with ΔT = the freezing point depression = 2.90 °C
⇒with i = the van't Hoff factor of potassium bromide = 1.9
⇒with Kf = the freezing point depression constant = 1.28 °C/m
⇒with m = the molality = TO BE DETERMINED
2.90 °C = 1.9 * 1.28 °C/m * m
m = 1.19 molal
Step 6: Calculate moles potassium bromide
molality = moles KBr / mass X
1.19 molal = moles KBr / 0.950 kg
moles KBr = 1.19 * 0.950 kg
Moles KBr = 1.13 moles KBr
Step 7: Calculate mass KBr
Mass KBr = moles KBr * molar mass KBr
Mass KBr = 1.13 moles * 119.0 g/mol
Mass KBr = 135.5 grams
We need 135.5 grams of potassium bromide