The solution contains 20 g of KI in 100 g of solution. The mass of water is therefore \( \text{Mass of water} = 100 \, \text{g} - 20 \, \text{g} = 80 \, \text{g} = 0.08 \, \text{kg} \). The moles of KI are calculated as \( \text{Moles of KI} = \frac{20 \, \text{g}}{166 \, \text{g/mol}} = 0.1205 \, \text{mol} \). Using the molality formula \( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \), the molality \( m \) is \( m = \frac{0.1205 \, \text{mol}}{0.08 \, \text{kg}} = 1.5 \, \text{mol kg}^{-1} \). Final Answer: \[\boxed{1.5 \, \text{mol kg}^{-1}}\]