An aqueous solution is made by dissolving 29.4 grams of aluminum acetate in 433 grams of water. The molality of aluminum acetate in the solution is

0.333 m

Explanation:

Molality (m) is moles of solute over kilograms of solvent.

Convert grams of the solute (aluminum acetate) to moles.

(29.4 g)/(204.11 g/mol) = 0.144 mol

Convert grams of the solvent (water) to kilograms.

433 g = 0.433 kg

Divide the solute by the solvent.

(0.144 mol)/(0.433 kg) = 0.333 m

The molality of the solution is 0.333 m.

it's filtration, but this is an assumption so my apologies if i got this

po-215 i got it right on the quiz