Question

when 61.6 g of alanine (C,H,NO,) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2.9 °C lower than the freezing point of pure X. On the other hand, when 61.6 g of ammonium chloride (NH,CI) are dissolved in the same mass of X, the freezing point of the solution is 7.3 °C lower than the freezing point of pure X Calculate the van't Hoff factor for ammonium chloride in X Be sure your answer has a unit symbol, it necessary, and is rounded to the correct number of significant digits.

Answer 1

Answer:

1.5

Explanation:

To solve this problem we use the freezing-point depression equation:

• ΔT = Kf * m * i

Where ΔT is the difference in temperature, Kf is the cryoscopic constant of the solvent, m is the molality, and i is the van't Hoff factor.

For the case of alanine (C₃H₇NO₂, MW = 89 g/mol):

• ΔT = 2.9 °C

m = mol alanine / kg X

1150 g X ⇒ 1150 / 1000 = 1.15 kg

61.6 g alanine ÷ 89g/mol = 0.692 mol

• m = 0.692 mol / 1.15 kg = 0.601 m

• i = 1

So now we calculate the cryoscopic constant of the liquid X:

• ΔT = Kf * m * i

• 2.9 °C = Kf * 0.601 m * 1

• Kf = 4.82 °C / m

Now we use the same formula for the case of ammonium chloride (NH₄Cl,  MW = 53.49 g/mol)

• ΔT = 7.3 °C

61.6 g NH₄Cl ÷ 53.49 g/mol = 1.15 mol

• m = 1.15 mol / 1.15 kg = 1 m

And calculate i :

• 7.3 °C = 4.82 °C/m * 1m * i

• i = 1.5

The van't Hoff factor is unitless.