Question

When 89.1 g of urea CH₄N₂O are dissolved in 950. g of a certain mystery liquid x, the freezing point of the solution is 6.7°C lower than the freezing point of pure x. on the other hand, when 89.1 g of ammonium chloride NH₄Cl are dissolved in the same mass of x, the freezing point of the solution is 13.2°C lower than the freezing point of pure x. calculate the van’t hoff factor for ammonium chloride in x.

Answer 1

To calculate the van't Hoff factor for ammonium chloride in the mystery liquid x, we can use the formula ΔTf1 = Kf × m1 × i1. Using the given freezing point depressions and the molality, we can solve for the van't Hoff factor,

Step-by-step explanation:

To calculate the van't Hoff factor for ammonium chloride in a certain mystery liquid x, we can use the formula:

ΔTf1 = Kf × m1 × i1

Where ΔTf1 is the freezing point depression, Kf is the molal freezing-point depression constant, m1 is the molality of the solute, and i1 is the van't Hoff factor. We are given that the freezing point depression for urea is 6.7°C and for ammonium chloride is 13.2°C. Let x be the van't Hoff factor for ammonium chloride, we can set up the following equation:

-6.7 = Kf × 0.0891 / 0.950 × x

-13.2 = Kf × 0.0891 / 0.950 × 2

Solving these equations, we find that x is approximately 1.55.