The given question is incomplete, the complete question is:
When 238.g of benzamide (C7H7NO) are dissolved in 600.g of a certain mystery liquid X, the freezing point of the solution is 5.0°C lower than the freezing point of pure X. On the other hand, when 238.g of iron(III) chloride are dissolved in the same mass of X, the freezing point of the solution is 11.5°C lower than the freezing point of pure X.
Calculate the van't Hoff factor for iron(III) chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits.
Answer:
The correct answer is 3.0.
Step-by-step explanation:
Based on the given question, the weight of benzamide given is 238 grams, the molecular mass of benzamide is 121.14 gram per mole. The benzamide is dissolved in 600 grams of liquid X, therefore, the solution's total weight is,
= 238 + 600 = 838 grams
In case of benzamide, ΔTf = kf × molality (It is given that the solution's freezing point is 5 degree C lesser in comparison to the pure X's freezing point). Now putting the values we get,
5 = kf × 238 / (121.14×838) × 1 (Van't Hoff factor) -------- (i)
In case of benzamide, the van't Hoff factor will be 1, as it is neither associate nor dissociate.
On the other hand, the mass of ferric chloride given is 238 grams getting dissolved in the similar mass of X, therefore again the total mass of the solution will be 838 grams. The freezing point of the solution is 11.5 degree C lesser than the pure X's freezing point. The molecular mass of ferric chloride is 162.2 gram per mol.
For FeCl3,
ΔTf = kf × molality
11.5 = kf × 238/ (162.2 × 838) × i (Van't Hoff factor)------ (ii)
Now dividing equation (ii) by (i) we get,
11.5 / 5 = (kf/kf) × (121.14 / 162.2) × i
i = 3.0