Question

20 L of a 3.5 molar aqueous solution of acedic acid is added to 23 L of methanol. What are the mole fractions of each species?

Answer 1

Ans :

Molarity = number of moles of solute / volume of solution in L

3.5 = n / 20

n = 20 x 3.5

= 70 mol acetic acid

Density of methanol is 792 g/L

Density = mass / volume

792 = m / 23

mass = 18216 g

Molar mass of methanol is 32.04 g/mol

So number of moles = 18216 / 32.04 = 568.54 mol methanol

Total number of mol = 70 mol + 568.54 mol = 638.54 mol

Mole fraction = number of mol / total number of mol

Mole fraction of acetic acid = 70 / 638.54 =0.110

Mole fraction of methanol = 568.54 / 638.54 = 0.890