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If a 0.610 m aqueous solution freezes at –3.00 °c, what is the van\'t hoff factor, i, of the solute? kf values can be found here.

Respuesta :

Dejanras
ΔT = 3°C = 3 K.
Kf(H₂O) = 1,853 K·kg/mol.
b = 0,610 m = 0,610 mol/kg.
b = n ÷ m(H₂O).
ΔT = Kf · b · i.
i = ΔT / Kf · b.
i = 3 K / 1,853 K·kg/mol · 0,610 mol/kg.
i = 2,65.
i - Van 't Hoff factor.
b - molality.
ΔT - change in temperature.
pstnonsonjoku

The Van't Hoff factor of the solution is 2.6.

We know that;

ΔT = k m i

ΔT = freezing point depression

K = freezing constant

m = molaitity of the solution

i = Van't Hoff factor

Now, to find the Van' Hoff factor,

i = 0 -(–3.00 °c)/1.86 oC m-1 × 0.610 m

i = 2.6

The Van't Hoff factor of the solution is 2.6.

Learn more about Van't Hoff factor: https://brainly.com/question/3930479

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