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.

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).