**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