What is molality of 20% NaOH (m/m)?

20 g of NaOH is present in 100 g of solution.

Mass of solvent = 100 - 20 = 80 g

Mol. mass of NaOH = 40 g/mol

No of moles = \(\frac{20}{40}\) = 0.5

Molality = \(\frac{0.5}{80}\)x1000 = 6.25 mol/kg