If a 0.710 m 0.710 m aqueous solution freezes at − 2.00 ∘ C, −2.00 ∘C, what is the van't Hoff factor, i , i, of the solute?

The van't Hoff factor of the solute is 1.51

Explanation:

Step 1: Data given

Molality = 0.710 molal

The aqueous solution freezes at − 2.00°C

Freezing point depression constant of water = 1.86 °C/m

Step 2: Calculate the van't Hoff factor

ΔT = i*Kf * m

⇒ with ΔT = The difference between the feezing point of pure and solution = 2.00°C

⇒ i the van't Hoff factor = TO BE DETERMINED

⇒ Kf = Freezing point depression constant of water = 1.86 °C/m

⇒ m = the molality of the solution = 0.710 molal

2.00 = i * 1.86 * 0.710

i = 1.51

The van't Hoff factor of the solute is 1.51