Question

A solution is prepared by dissolving 28.4 g acetone in 238 g of water. The final volume of the solution is 247mL. For the solution, calculate the concentration of each unit.

a.molarity

b.mass percent

c.molality

d.mole fraction

Answer 1

Number of moles of acetone = mass / molar mass

= 28.4 g / 58.08 gmol-1

= 0.488 mol

Similarly, Number of moles of water = 238 g / 18 gmol-1

= 13.2 mol

a) molarity = number of moles of solute / volume of solution in litres

= 0.488 mol / 0.247 L

= 1.975 M

b) mass percent = (mass of solute / mass of solution )* 100

= ( 28.4 / (28.4 + 238) ) * 100

= 10.6 %

c) molality = moles of solute / mass of solvent in kg

= 0.488 mol / 0.238 g

= 2.05 m

d) mole fraction of acetone = moles of acetone / mol of (acetone + water)

= 0.488 mol /( 0.488 + 13.2) mol

= 0.035

Mol fraction of water = 1- 0.035

= 0.965