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

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error