Assuming ideal behavior, what is the freezing point of a 0.030 m solution of al2(so4)3 ?

The freezing point of ideal solutions lower respect the freezing point of the pure solvent following the freezing depression formula:

ΔT = i * Kf * m

Where:

-  i is the vant'Hoff, which is the number of partilces (ions in this case)  that are the solute forms when dissociate.

- kf is cryoscopic constant of the pure solvent (water because we assume it is an aqueous solution).

Then kf = - 1.86 °C / mol

- and m is the molality, m = 0.030 m

i = 2 + 3 = 5, given the one molecule of Al2 (SO4)3 produces 2 ions of Al(+) and 3 ions of SO4 (2-).

Then, ΔT = 5 * 1.86°C/m * 0.030 m = 0.279 °C.

That is the depression of the freezing point which you have to subtract from the freezing point of the pure water => 0°C - 0.279°C = - 0.279 °C.

- 0.279°C 

 

Idid the calculation and i get 875,000,000 which can be written as 8.75 x 10^6?

Alpha particles have positive charge, beta particles have negative charge, and answer 4 just doesn’t make sense. thus, you are left with only one possible answer.