Question

A solution of ethanol (C2H6O) in water is sometimes used as a disinfectant. 1.00 L of this solution contains 553 g of ethanol and 335 g of water. What is the molality of the ethanol in this solution?

Answer 1

The molality of a solution can be calculated by dividing the number of moles of solute (ethanol) by the mass of the solvent (water) in kilograms. To calculate the moles of ethanol, we need to know its molar mass:

C2H6O (ethanol) molar mass = 2(12.01 g/mol) + 6(1.01 g/mol) + 16.00 g/mol = 46.07 g/mol

Next, divide the total mass of ethanol (553 g) by its molar mass (46.07 g/mol) to get the number of moles:

553 g / 46.07 g/mol = 12.0 mol

Next, convert the mass of water (335 g) to kilograms:

335 g / 1000 g/kg = 0.335 kg

Finally, divide the number of moles of ethanol (12.0 mol) by the mass of water in kilograms (0.335 kg) to get the molality:

12.0 mol / 0.335 kg = 35.8 mol/kg (to 3 significant figures)