When 61.6 g of alanine (C3H7NO2) 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 (NH4CI) 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.

Explanation:

From the given information:

TO start with the molarity of the solution:

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is

Using the depression in freezing point, the molar depression constant of the solvent

The freezing point of the solution

The molality of the solution is:

Molar depression constant of solvent X,

Hence, using the elevation in boiling point;

the Vant'Hoff factor

i believe it is c.) 1575 j.

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