What would be the freezing point of a 1.50 m solution of NaNO₃ in water? (K_f = 1.86 $\degree$C)

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Quizplus · Accepted Answer

The freezing point depression formula is ΔTf = iKf m, where ΔTf is the change in freezing point, i is the van't Hoff factor (number of particles the solute dissociates into), Kf is the freezing point depression constant (1.86 °C/m for water), and m is the molality of the solution. NaNO₃ appears to be a typographical error, but assuming it represents a salt that dissociates into two particles (Na+ and a generic anion), i = 2. Thus, ΔTf = 2 * 1.86 °C/m * 1.50 m = 5.58 °C. Since this is a depression, the freezing point is -5.58 °C.

What Would Be the Freezing Point of a 1 °\degree° C) A)5

What Would Be the Freezing Point of a 1 $\degree$ C)
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