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Calculate the `(a)` molality, `(b)` molarity, and `(c)` mole fraction of `KI` if the density of `20% (` mass `//` mass `)` aqueous `KI` is `1.202 g m L^(-1)`.

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20% aq. KOH solution `implies20g` of KI in 100g
solution
`therefore` Mass of solvent `=100-20=80g`
(i). Molality `=("no. of moles of KI")/("mass of solvent (kg)")`
`=(0.120)/(0.080)=1.5" mol "kg^(-1)`
(ii). Density of solution `=1.202" g "mL^(-1)`
Volume of solution `=(100)/(1.202)=83.2mL`
`=0.0832L`
`therefore`Molarity `=(0.120)/(0.0832)=1.44M`
(iii). No. of moles of KI `=0.120`
`n_(H_(2)O)=(80)/(18)=4.44`
`x_(KI)=(0.120)/(0.120+4.44)`
`=(0.120)/(4.560)=0.0263`.

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