Calculate:

$A$: Molality of $KI$ if the density of $20\%\left(\text{mass/mass}\right)$ aqueous $KI$ is $1.202gm{L}^{-1}$.

$B$: Molarity of $KI$ if the density of $20\%\left(\text{mass/mass}\right)$ aqueous $KI$ is $1.202gm{L}^{-1}$.

$C$: Mole Fraction of $KI$ if the density of $20\%\left(\text{mass/mass}\right)$ aqueous $KI$ is $1.202gm{L}^{-1}$.

$A$:

Given:

$20\%$ (mass/mass) aqueous solution of $KI$.
This means $20g$ of $KI$ is dissolved in $80g$ of water.
So, mass of solute $W_2=20g$ and mass of solvent $W_1=80$
Molar mass of solute $\left(KI\right)$
$=39+127$
$=166gmo{l}^{-1}$

Molality of $KI$:

We know, $\text{molality}\left(m\right)=\frac{\text{mass of solute}\left(g\right)}{\text{molar mass ofsolute}\times \text{mass of solvent(kg)}}$

Density of the aqueous $KI=1.202gm{L}^{-1}$

$m=\frac{20\times 1000}{166\times 80}=\frac{250}{166}=1.5m$

Final Answer:
The molality of $KI$ is $1.5m$.

$B$:

Given:
Mass percentage$=20\%$
And density of solution $=1.202gm{L}^{-1}$

Molarity of $KI$
We know, molarity $\left(M\right)=\frac{\text{mass percentage}\times \text{density}\times 10}{\text{molar mass}}$

Molarity$\left(M\right)=\frac{20\times 1.202\times 10}{166}=1.45M$

Final Answer:
The molarity of $KI$ is $1.45M$.

$C$:

Given:
$20\%$ (mass/mass) aqueous solution of $KI$.
This means $20g\text{of}KI$ is dissolved in $80g$ of water.
So, mass of solute $\left(W_2\right)=20g\text{and mass of solvent}\left(W_1\right)=80g$
Molar mass of solute $\left(KI\right)$
$=39+127$
$=166gmo{l}^{-1}$

Moles of compound

Number of moles of $KI,$

$n_{KI}=\frac{\text{Mass of KI}}{\text{Molar mass of KI}}=\frac{20}{166}=0.12mol$

Number of moles of $H_{2}O,$

$n_{H_{2}O}=\frac{\text{Mass of}{H}_{2}O}{\text{Molar mass of}{H}_{2}O}=\frac{80}{18}=4.44mol$

Mole fraction of compounds

Now, mole fraction of $\left(KI\right)$,

${\chi }_{KI}=\frac{n_{KI}}{n_{KI}+n_{H_{2}O}}$

${\chi }_{KI}=\frac{0.12}{0.12+4.44}$
$\left({\chi }_{KI}\right)=\frac{0.12}{4.56}=0.0263$

Final Answer:
The mole fraction of $KI$ in the given solution is $0.0263$.