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The given question is incomplete. The complete question is:
When 282. g of glycine (C2H5NO2) are dissolved in 950. g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 282. g of ammonium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for ammonium chloride in X.
Answer: the van't Hoff factor for ammonium chloride is 1.74
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
= freezing point constant = ?
i = 1 ( for non electrolyte)
m= molality
Weight of solvent (X)= 950 g = 0.95 kg
Molar mass of solute (glycine) = 75.07 g/mol
Mass of solute (glycine) = 282 g
ii)
Weight of solvent (X)= 950 g = 0.95 kg
Molar mass of solute (ammonium chloride) = 53.49 g/mol
Mass of solute (ammonium chloride) = 282 g
Thus the van't Hoff factor for ammonium chloride is 1.74