Calculate the molality of a 24.4% (by mass) aqueous solution of phosphoric acid (h3po4).

Molality pertains to a concentration of a mixture expressed in moles of solute per kg of solvent. Based on the given problem, the solute is the phosphoric acid with molecular weight of 97.99 g/mol. Assuming 1 g of sample, the mole of solute is equal to 0.244 * (97.99) = 2.49x10-3 moles. The remaining percentage, t that is, 1-0.244 is the amount of solvent, which is equal to 0.756 g based on 1 gram assumption. Therefore, the molality is,

(0.244) * (mol/97.99 g) / (1-0.244) * (1 kg/1000 g) = 3.29 molala