High School

The freezing point of an aqueous solution of MgBr2 was found to be -5.7 °C. Calculate the boiling point of an aqueous solution of AlCl3 that has the same molality as the MgBr2 solution.

Given:
- \( K_f \) of water = 1.86 °C/m
- \( K_b \) of water = 0.512 °C/m

Options:
A. 102.1 °C
B. 101.1 °C
C. 101.5 °C
D. 99.9 °C
E. 100.5 °C

Answer :

The boiling point of an AlCl3 solution with the same molality as the given MgBr2 solution, considering the full dissociation of both ionic compounds in water, would be 101.536 °C. Therefore, amongst the provided options, the closest answer is (B) 101.5 °C.

The subject matter pertains to colligative properties, which are properties of a solution that depend only on the number, not the identity, of the solute particles. The key information given here is the freezing point depression of a MgBr2 solution and we're asked to find the boiling point elevation of a solution of AlCl3 with the same molality. Both MgBr2 and AlCl3 are ionic compounds and will dissociate in water. The change in boiling point ΔTb is calculated by the formula ΔTb = Kb * m * i. Where 'Kb' is the boiling point elevation constant for water (0.512 °C/m), 'm' is the molality of the solution, and 'i' is the van 't Hoff factor.

MgBr2 and AlCl3 both fully dissociate into three ions in solution (Mg2+ + 2Br-, and Al3+ + 3Cl-). Thus, the 'i' value for both is 3. Since both solutions have the same molality, the change in boiling point would be the same for both. ΔTb for this solution is then 0.512 * 3 = 1.536 °C. Thus, the boiling point of the AlCl3 solution would be 100 °C (normal boiling point of water) + 1.536 °C = 101.536 °C. Among the given options, 101.5 °C is the closest one.

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