Question

A 95.0 wt% solution of ethanol, C2H5OH (mw = 46.07) in water has a density of 0.804 g/ml. Find the mass of 1.00 L of this solution and the number of grams of ethanol per liter. What is the molar concentration of ethanol in this solution? Find the molality of ethanol in this solution, considering H2O to be the solvent (even though H2O is really the solute in this case).

Answer 1

We have 95.0 wt% ethanol solution which simply means that 100.0 g of the solution has 95.0 g ethanol.

Density of the solution is 0.804 g/mL; therefore, mass of 1.0 L of the solution = (1.0 L)*(1000 mL/1 L)*(0.804 g/mL) = 804.0 g (ans).

Mass of ethanol in 1.0 L solution = (804.0 g solution)*(95.0 g ethanol/100 g solution) = 763.8 g ethanol (ans).

Density of the solution is 0.804 g/mL; therefore, volume of the solution is (mass of solution)/(density of solution) = (100.0 g)/(0.804 g/mL) = 124.378 mL.

M.W. of ethanol is 46.07; therefore, mole(s) of ethanol corresponding to 95.0 g ethanol = (mass of ethanol)/(M.W. of ethanol in g/mol) = (95.0 g)/(46.07 g/mol) = 2.062 mole.

Molar concentration of ethanol in the solution = (moles of ethanol)/(volume of solution in L) = (2.062 mole)/[(124.378 mL)*(1 L/1000 mL)] = 16.57849 mol/L ≈ 16.578 M (ans).

The solution is made up of ethanol in water. 100.0 g of the solution contains 95.0 g ethanol; therefore, mass of water in the solution = (100.0 – 95.0) g = 5.0 g = (5.0 g)*(1 kg/1000 g) = 0.005 kg.

Again, 95.0 g ethanol = 2.062 mole ethanol.

Molality of ethanol = (moles of ethanol)/(kg of water) = (2.062 mole)/(0.005 kg) = 412.4 mole/kg = 412.4 m (ans).