If a 0.900 𝑚 aqueous solution freezes at −2.10 °C , what is the van't Hoff factor, 𝑖 , of the solute?

Question

i=?

Answer 1

Where to start:

1. Identify the given:
2. molality m = 0.900 m
3. freezing point of solution T_soln = -2.10 °C
4. What you're looking for:
5. van't hoff factor i =?
6. Identify which formula has these variables
7. ΔT = -K_f · m · i
8. ΔT = T_soln - T_H2O
9. so we can rewrite the equation as T_soln - T_H2O = -K_f · m · i

T_soln = freezing point of the solution (unit: °C)

T_H2O = freezing point of water (0 °C)

K_f = freezing point of depression constant (typically given Kf of water = -1.86 °C/m)

m = molality (moles solute/kg solvent) (unit: m)

i = van't hoff factor (number of ions) (no units)

1. Plug in values and isolate 'i'

-2.10 = -1.86 × 0.900 × i

-2.10 / (-1.86× 0.900) = i

1.25 = i

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