A 5.431 m formic acid (HCO2H) solution in water has a density of 1.047 g/mL. Find the molarity of the solution and the percent by mass and mole fraction of formic acid

Answer 1

5.431 m = 5.431 moles HCO2H / kg solvent. If we assume we used 1 kg H2O, then we have...

5.431 moles HCO2H x 46 g / mol = 250 g formic acid + 1000 g H2O

250 g + 1000 g = 1250 g solution

1250 g soln x 1 ml/1.047 g = 1194 ml = 1.194 L

Molarity (M) = moles formic acid / L = 5.431 mol / 1.194 L = 4.549 M

Percent by mass = 250 g / 1250 g (x100%) = 20% by mass

mole fraction formic acid:

moles formic acid = 5.431

moles H2O = 1000 g x 1 mol / 18 g = 55.6666.... mol

total moles = 55.66.. + 5.431 = 60.987

mole fraction formic acid = 5.431 / 60.987 = 0.08905 = mole fraction