[CripplingStudyhabbits]

Water has a Kf of 1.86 °C/m. Calculate the new freezing point of an aqueous solution made by mixing 34.88 g water and 1.420 g MgCl2.

Ok so freezing point depression, ΔT= Kf x m x i

I took the grams of MgCl₂ 1.420g, then divided it by its molar mass (95.211) = (.0149)
Grams of water 34.88, then converted to Kg, = (.03488)

molality is, molarity(solute)/kg(solvent), .0149/.03488=.427

Kf=1.86

The Van't Hoff Factor of MgCl₂ is 3

.427 x 1.86 x 3 = 2.383

Then 0°C - 2.383 = -2.383°C

but it says it is wrong 

somebody help, HW is worth 30% of my grade

[AWK]

The Van't Hoff Factor of MgCl2 is 3

Real (measured) factor is 2.7

[Borek]

Are you sure the problem is not with sig figs?

Technically AWK is right, but whether you are expected to use experimentally determined value that takes into account activity coefficients, creation of ionic pairs and whatnot, depends on the course level. For 101 3 would be most likely right.

[CripplingStudyhabbits]

Are you sure the problem is not with sig figs?

The question says to enter exactly 4 sig figs

[Borek]

I wonder if it is not a case of rounding errors. Molality - while close to 0.427 - is not exactly that, so in the end I got a slightly different number as an answer.

[CripplingStudyhabbits]

I wonder if it is not a case of rounding errors. Molality - while close to 0.427 - is not exactly that, so in the end I got a slightly different number as an answer.

thank you I shall try again