Given : Let mass of solution in water = $100\ \text{g},$ then mass of $\mathrm{KI} = 20\ \text{g}$ $\therefore\;$ Mass of solvent (water) $100-20$ $=80\ \text{g}=0.080\ \text{kg}$(a) Calculation of molalityMolar mass of $\mathrm{KI}=39+127$ $=166\ \text{g}\ \text{mol}^{-1}$ $\therefore\;$ Moles of $\mathrm{KI}=\frac{20\ \text{g}}{166\ \text{g}\ \text{mol}^{-1}}$ $=0.120$Molality of solution $=\frac{\text{Number of moles of }\mathrm{KI}}{\text{Mass of solvent in }\text{kg}}$ $=\frac{0.120\ \text{mol}}{0.080\ \text{kg}}=1.5\ \text{mol}\ \text{kg}^{-1}$(b) Calculation of molarityDensity of solution = $1.202\ \text{g}\ \text{mL}^{-1}$ $\therefore$ Volume of solution $=\frac{100\ \text{g}}{1.202\ \text{g}\ \text{mL}^{-1}}$ $=83.2\ \text{mL}=0.0832\ \text{L}$Molarity of solution $=\frac{\text{Number of moles of solute}}{\text{Volume of solution in }\text{L}}$ $=\frac{0.120\ \text{mole}}{0.0832\ \text{L}}=1.44\ \text{M}$(c) Calculation of mole fraction of KINo. of moles of KI = 0.120No. of moles of water $=\frac{\text{Mass of water}}{\text{Molar mass of water}}$ $=\frac{80\ \text{g}}{18\ \text{g}\ \text{mol}^{-1}}=4.44$Mole fraction of $=\frac{0.120}{0.120+4.44}=\frac{0.120}{4.560}=0.0263$