(i) Molar mass of KI

= 39 + 127

= 166 g mol^-1

20% (mass/mass) aqueous solution of KI means 20g of KI is present in 100 g of solution.

That is, 20 g of KI is present in (100 - 20) g of water

= 80 g of water

Therefore, molality of the solution = \(\frac{Moles\;of\;KI}{Mass\;of\;water\;in\;kg}\)

= \(\frac{\frac{20}{166}}{0.08}\)m

= 1.506 m

= 1.51 m (approximately)

(ii) It is given that the density of the solution

= 1.202 g mL^-1

∴ Volume of 100 g solution = \(\frac{Mass}{Density}\)

= \(\frac{100g}{1.202\;g\;mL^{-1}}\)

= 83.19 mL

= 83.19 × 10^-3L

Therefore, molarity of the solution

= \(\frac{\frac{20}{166}mol}{83.19\times 10^{-3}L}\)

= 1.45 M

(iii) Moles of KI = \(\frac{20}{166}\) = 0.12 mol

Moles of water = \(\frac{80}{18}\) = 4.44 mol

Therefore, mole fraction of KI

= \(\frac{Moles\;of\;KI}{Moles\;of\;KI + Moles\;of\;water}\)

= \(\frac{0.12}{0.12 + 4.44}\)

= 0.0263