Question 7 When of benzamide are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing point of pure . On the other hand, when of iron(III) chloride are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for iron(III) chloride in . Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.

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Explanation:

moles of benzamide = mass / Molar mass of it = 70.4g / ( 121.14g/mol) = 0.58 mol

Molality = moles of solute ( benzamide) / ( solvent mass in kg) = 0.58 mol / ( 0.85kg) = 0.6837

we have formula dT = i x Kf x m , where dT = change in freezing point = 2.7C , i = vantoff factor = 1 for non dissociable solutes , Kf = freezing oint constant of solvent , m = 0.6837

hence 2.7C = 1 x Kf x 0.6837m

Kf = 3.949 C/m

we use this Kf value for calculating i for NH4Cl , where moles of NH4Cl = ( 70.4g/53.491g/mol) =1.316 mol

molality = ( 1.316mol) / ( 0.85kg) = 1.5484 , dT = 9.9

hence 9.9 = i x 3.949C/m x 1.5484 m

i = 1.62

po-215 i got it right on the quiz