What are the molality and mole fraction of solute in a 34.6 percent by mass aqueous solution of formic acid (HCOOH)?

Question

What are the molality and mole fraction of solute in a 34.6 percent by mass aqueous solution of formic acid (HCOOH)?

Answer 1

Let us assume that we have 1000 g of solution.

The mass of formic acid = 34.6% x 1000 g = 346 g formic acid

The mass of H2O will therefore be 1000 g - 346 g = 654 g H2O

Moles of formic acid = 346 g x 1 mol/46 g = 7.52 moles formic acid

Moles H2O = 654 g x 1 mol/18 g = 36.3 moles H2O

Molality of formic acid = moles/kg = 7.52 mol / 0.654 kg = 11. 5 m

Mole fraction of formic acid = mol formic acid/total moles = 7.52 mol / 7.52 + 36.3 = 7.52/43.8 = 0.172