Question

If a 0.760 m aqueous solution freezes at − 3.60 ∘ C, what is the van't Hoff factor, i , of the solute?

Answer 1

The van't Hoff factor is 2.55

Step-by-step explanation:

Step 1: Data given

molality = 0.760 molal

The aqueous solution freezes at -3.60 °C

Kb of water = 1.86°C/m

Step 2: Calculate the van't Hoff factor

ΔT = i*Kb*m

3.60 = i * 1.86 * 0.760

i = 3.60( 1.86*0.760)

i = 2.55

The van't Hoff factor is 2.55

Answer 2

2.556.

Step-by-step explanation:

Freezing point depression is defined as the decrease in temperature when a solute is added to a solvent.

The resulting liquid solution has a lower freezing point than the pure solvent because the chemical potential of the solvent in the mixture is lower than that of the pure solvent, the difference between the two being proportional to the natural logarithm of the mole fraction.

Mathematically,

ΔTf = Kf * b * i

where:

ΔTemp.f = the freezing-point depression. It is defined as Temp.f1 (pure solvent) − Temp.f2 (solution).

Kf = the cryoscopic constant, which is dependent on the properties of solvent.

b = the molality (moles solute per kilogram of solvent)

i = van 't Hoff factor

Kf = 1.853 K·kg/mol

Temp.f1 = 0°C

Temp.f2 = -3.60 °C

b = 0.76 m

ΔTf = Kf * b * i

0 - (-3.6) = 1.853 * 0.76 *i

i = 3.6/1.4083

= 2.556