What are the molality and mole fraction of solute in a 36.6 percent by mass aqueous solution of formic acid (HCOOH)?

Question

What are the molality and mole fraction of solute in a 36.6 percent by mass aqueous solution of formic acid (HCOOH)?

Answer 1

36.6 percent by mass of formic acid means 36.6 g of formic acid per 100 g of solution. This can be translated to be 36.6 g formic acid + 63.4 g of water.

Molality = moles solute/kg solvent

moles of solute = 36.6 g HCOOH x 1 mol HCOOH/46 g = 0.796 moles HCOOH

kg of solvent = 63.4 g x 1 kg/1000 g = 0.0634 kg

Molality = 0.796 moles/0.0634 kg = 1.25 m

To find mole fraction, we need to first find moles of H2O:

63.4 g H2O x 1 mol/18 g = 3.52 moles

We add this to the moles of HCOOH (0.796 moles) to get a total of 4.32 moles.

Mole fraction of HCOOH is moles HCOOH/total moles = 0.796 moles/4.32 moles = 0.184 <-- mole fraction