Question

What are the molality and mole fraction of solute in a 22.2 percent by mass aqueous solution of formic acid (HCOOH)?

Answer 1

Since there is no sample, let us assume 100 g of the solution:
(22.2% of 100 g) / (46.0254 g HCOOH/mol) = 0.48234 mol HCOOH
(100 g - 22.2 g) = 77.8 g = 0.0778 kg water
(0.48234 mol HCOOH) / (0.0778 kg) = 6.1997 mol/kg = 6.20 m HCOOH
(77.8 g H2O) / (18.01532 g H2O/mol) = 4.3185 mol H2O
(0.48234 mol HCOOH) / (0.48234 mol + 4.3185 mol) = 0.100 [the mole fraction of HCOOH]