Question

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

Answer 1

molality = (W /MW) (1000 / mass of solvent in g)

assume mass of solvent = 100 g

5.431 = (W / 46.03) (1000 / 100)

W / 46.03 = 0.5431

W = 25.0 g  

mass of solution = mass of solvent + mass of solute

mass of solution = 100 + 25.0 = 125 g

density = mass / volume

volume = mass / density

volume of solution = 125 / 1.047

volume of solution = 119.4 mL

now

molarity = (W /MW) (1000 / volume of solution in mL)

molarity = (25 / 46.03) (1000 / 119.4)

molarity = 4.55 M

2) % mass = (mass of solute / mass of solution) x 100

% mass = (25 / 125) x 100

% mass = 20 %

3) mole fraction of solute = moles of solute / total moles

moles of acid = 25 / 46.03 = 0.543

moles of water = 100 / 18 = 5.55

total moles = 5.55 + 0.543 = 6.093

mole fraction = 0.543 / 6.093

mole fraction = 0.089