Calculate $(a)$ molality,$(b)$ molarity,and $(c)$ mole fraction of $KI$ if the density of $20 \%$ (mass/mass) aqueous $KI$ solution is $1.202 \, g \, mL^{-1}$.

(N/A) Molar mass of $KI = 39 + 127 = 166 \, g \, mol^{-1}$.
$20 \%$ (mass/mass) aqueous solution of $KI$ means $20 \, g$ of $KI$ is present in $100 \, g$ of solution. Thus,$20 \, g$ of $KI$ is present in $(100 - 20) \, g = 80 \, g$ of water.
Molality $(m) = \frac{\text{Moles of } KI}{\text{Mass of water in } kg} = \frac{20 / 166}{0.080} \, m = 1.506 \, m \approx 1.51 \, m$.
$(b)$ Density of solution $= 1.202 \, g \, mL^{-1}$.
Volume of $100 \, g$ solution $= \frac{\text{Mass}}{\text{Density}} = \frac{100}{1.202} \, mL = 83.19 \, mL = 0.08319 \, L$.
Molarity $(M) = \frac{\text{Moles of } KI}{\text{Volume of solution in } L} = \frac{20 / 166}{0.08319} \, M = 1.448 \, M \approx 1.45 \, M$.
$(c)$ Moles of $KI = \frac{20}{166} = 0.1205 \, mol$.
Moles of water $= \frac{80}{18} = 4.444 \, mol$.
Mole fraction of $KI = \frac{n_{KI}}{n_{KI} + n_{H_2O}} = \frac{0.1205}{0.1205 + 4.444} = \frac{0.1205}{4.5645} = 0.0264$.