Question

An aqueous solution has a mole fraction of solute of 0.0385. What is the molality of this solution?

(I know the answer is 2.222 m I just need help with the steps to get there)

Answer 1

not Molfraction of solute, lobte = — Polite+ PHO = 0.0385 -0.0385 1 0.0385 mol 0.0385 mol + 0.9615 mol No of moles of solute, ale = 0.0385 mol No of moles of water (H20), %4,0 = 0.9615 mol 18.02 g mass of 0.800 moles of H2,0 = 0.9615 mol H,Ox; 1 mol H20 = 17.33 g = 0.01733 kg _ no of moles of solute Molality of solution = - mass of solvent in kg 0.0385 mol = 2.222 mol/kg = 2.222 m 0.01733 kg