What is the mole fraction of solute and the molal concentration for an aqueous solution that is 20.0% NaOH by mass?

Let's assume we have 100 grams of this solution, and therefore 20.0 grams of NaOH, and 80.0 grams of water.

NaOH has a molar mass of 40.00 grams/mole, so we can convert NaOH to moles: (20.0 grams of NaOH) * (1 mole NaOH/40.00 grams NaOH) = 0.500 moles NaOH.

Next, we have a molar mass of water of 18.02 g/mol, so we can convert water to moles:

80.0 grams H2O * (1 mole H2O/18.02 grams) = 4.44 moles H2O

The mole fraction of NaOH is the moles of NaOH over the total moles of all parts of the solution. Therefore:

(0.500 moles NaOH) / (0.500 moles + 4.44 moles) = 0.101

0.101 = mole fraction of solute

The molal concentration is defined as the number of moles of solute over the number of kilograms of solvent (water).

We have 0.500 moles of NaOH, and 0.0800 kg of water, so it becomes:

(0.500 moles NaOH) / (0.0800 kg H2O solvent) = 6.25 molal solution