An aqueous solution has a boiling point of 102.45°C. What is its freezing point? ( Kf = 1.86°C m-1 and Kb = 0.51°C m-1 )

The freezing point for the solution is -8.9°C

Explanation:

We must combine the two colligative properties of freezing point depression and boiling point elevation to solve this excersise.

ΔT = Kb . m

ΔT = T° boiling point of solution - T° boiling of pure solvent

Kb = Ebullioscopic constant

m = molality

102.45°C - 100°C = 0.51°C/m . m

2.45°C / 0.51 m /°C = m → 4.80 mol/kg

We found out the molality in the boiling point elevation to replace at the freezing point depression formula

ΔT = Kf . m

ΔT = T° freezing pure solvent - T° freezing solution

Kf = Cryoscopic constant

0°C - T° freezing solution = 1.86°C/m . 4.80 m

T° freezing solution = - 8.9°C

the answer is b because, investigating whether another species of moth also drinks tears might be interesting, but it would not directly explain why this species of moth does.

to explain why this species of moth drinks tears, the scientist could ask: "what nutrients that moths need are in elephant tears? "

the scientist could experiment with moth repellent or oter ways to keep the moths away, but that would not answer his original question.