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.

1.5

Explanation:

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

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):

m = mol alanine / kg X

1150 g X ⇒ 1150 / 1000 = 1.15 kg

61.6 g alanine ÷ 89g/mol = 0.692 mol

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

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

61.6 g NH₄Cl ÷ 53.49 g/mol = 1.15 mol

And calculate i :

The van't Hoff factor is unitless.

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