At what temperature would a 1.15 m nacl solution freeze, given that the van't hoff factor for nacl is 1.9? kf for water is 1.86 ∘c/m .express your answer with the appropriate units.

-4.1°C

Explanation:

The freezing-point depression (ΔTf) is a colligative property that can be calculated using the following expression:

ΔTf = i × Kf × m

where,

i: van't Hoff factor

Kf: freezing-point depression constant

m: molality

ΔTf = 1.9 × 1.86°C/m × 1.15 m = 4.1 °C

The freezing-point of pure water is 0°C. The freezing-point of this solution is:

0°C - 4.1°C = -4.1°C

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