Question

A 0.89 m aqueous solution of an ionic compound with the formula mx has a freezing point of -3.0 ∘c . van't hoff factor?

Answer 1

Answer is: Van't Hoff factor (i) for this solution is 1,81.
Change in freezing point from pure solvent to solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
b -  molality, moles of solute per kilogram of solvent.
b = 0,89 m.
ΔT = 3°C = 3 K.
i = 3°C ÷ (1,86 °C/m · 0,89 m).

i = 1,81.