Answer:
The van't Hoff factor of NaCl in liquid X is 1.69
Explanation:
Step 1: Data given
Mass of urea = 78.6 grams
Molar mass of urea = 60.06 g/mol
Mass of liquid X = 700 grams = 0.700 kg
he freezing point of the solution is 4.9°C lower than the freezing point of pure X
When 78.6 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 8.5°C lower than the freezing point of pure X.
Molar mass of NaCl = 58.44 g/mol
Step 2: Calculate moles
Moles urea = mass / molar mass
Moles urea = 78.6 grams / 60.06 g/mol
Moles urea = 1.31 moles
Moles NaCl = 78.9 grams / 58.44 g/mol
Moles NaCl = 1.35 moles
Step 3: Calculate molality
Molality = moles / mass of liquid
Molality urea = 1.31 moles / 0.700 kg
Molality = 1.87 molal
Molality NaCl = 1.35 moles / 0.700 kg
Molality NaCl = 1.92 molal
Step 4: Calculate the freezing point depression constant of X
ΔT = i*Kf*m
⇒with ΔT = the freezing point depression = 4.9 °C
⇒with i = the van't hoff factor of urea = 1
⇒with Kf =the freezing point depression consant of X = TO BE DETERMINED
⇒with m = the molality of urea solution = 1.87 molal
4.9 °C = 1 * Kf * 1.87 molal
Kf == 4.9 / 1.87
Kf = 2.62 °C/m
Step 5: Calculate the van't Hoff facotr of NaCl in X
ΔT = i*Kf*m
⇒with ΔT = the freezing point depression = 8.5 °C
⇒with i = the van't hoff factor of urea = TO BE DETERMINED
⇒with Kf =the freezing point depression consant of X = 2.62 °C/m
⇒with m = the molality of urea solution = 1.92 molal
8.5 °C = i * 2.62 °C/m * 1.92 m
i = 8.5 / (2.62 * 1.92)
i = 1.69
The van't Hoff factor of NaCl in liquid X is 1.69
