When 177. g of alanine (C3H7NO2) are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is 5.9 °C lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is 7.2 °C lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X. Be sure your answer has a unit symbol, if necessary, and is rounded to the correct number of significant digits.

Van't Hoff factor of KBr is 1.63

Explanation:

Freezing point depression due the addition of a solute follows the formula:

ΔT = Kf×m×i

Where ΔT is change in freezing point, Kf is freezing point depression constant of the solvent X, m is molality of the solution (mol / kg) and i is Van't Hoff factor.

Moles of 177g of alanine (Molar mass: 89.09g/mol) are:

177g × (1mol / 89.09g) = 1.99 moles. In 0.800kg:

1.99mol / 0.800kg = 2.49m

Replacing in freezing point depression formula:

5.9°C = Kf×2.49m×1

Alinine has a Van't Hoff factor of 1

The Kf of the solvent is:

2.37 °C/m

Molality of the 177.0g of KBr solution (Molar mass: 119g/mol) is:

177.0g × (1mol / 119g) = 1.487 moles / 0.800kg = 1.859m

And the freezing point depression formula is:

7.2°C = 2.37°C/m×1.859m×i

1.63 = i

Van't Hoff factor of KBr is 1.63

what is the question?