(a) Molality of KI
(b) Molarity of KI
(c) Mole fraction of KI
(a) Molar mass of $\mathrm{Kl}=39+127=166 \mathrm{~g} \mathrm{~mol}^{-1}$
$20 \%$ aqueous solution of KI shows that $20 \mathrm{~g}$ of $\mathrm{Kl}$ is present in $100 \mathrm{~g}$ of solution.
i.e.,
$20 \mathrm{~g}$ of $\mathrm{Kl}$ is present in $(100-20) \mathrm{g}$ of water $=80 \mathrm{~g}$ of water
Hence, molality of the solution $=\frac{\text { Moles of KI }}{\text { Mass of water in } \mathrm{kg}}$
$=\frac{\frac{20}{166}}{0.08}m$
$=1.506~\text{m}$
$=1.51~\text{m (approx}\text{.) }$
(b) The destiny of the solution is given that $=1.202 g m L^{-1}$
Volume of $100 \mathrm{~g}$ solution $=\frac{\text { Mass }}{\text { Density }}$
$=\frac{100~\text{g}}{1.202~\text{g}~\text{m}{{\text{L}}^{-1}}}$
$=83.19~\text{mL}$
$=83.19 \times 10^{-3} L$
Hence, molarity of the solution $=\frac{\frac{20}{165} \mathrm{~mol}}{83.19 \times 10^{-3} L}$
$=1.45 \mathrm{M}$
(c) Moles of $\mathrm{KI}=\frac{20}{166}=0.12 \mathrm{~mol}$
Moles of water $=\frac{80}{18}=4.44 \mathrm{~mol}$
Hence, mole $=\frac{\text { Moles of } K I}{\text { Moles of } K I+\text { Moles of water }}$
Fraction of $\mathrm{KI}=\frac{0.12}{0.12+4.44}$ $=0.0263$