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

Explanation

`20%` (mass/mass) aqueous KI solution means that
Mass of KI = 20 g, `" "` Mass of solution in water = 100 g
`therefore" Mass of solvent (water) "=100-20=80g=0.080 kg`
(a) Calculation of molality
`"Molar mass of KI"=39+127="166 g mol"^(-1)" " therefore" Moles of KI"=("20 g")/("166 g mol"^(-1))=0.120`
`"Molality of solution"=("No. of moles of KI")/("Mass of solvent in kg")=("0.120 mole")/("0.080 kg")="1.5 mol kg"^(-1)`.
(b) Calculation of molarity
`"Density of solution "="1.202 g mL"^(-1)" "therefore" Volume of solution "=(100g)/("1.202 g mL"^(-1))="83.2 mL = 0.0832 L"`
`"Molarity of solution"=("Moles of solute")/("Volume of solution in L")=("0.120 mole")/("0.0832 L")="1.44 M".`
(c) Calculation of mole fraction of KI
No. of moles of KI = 0.120
`"No. of moles of water"=("Mass of water")/("Molar mass of water")=("80 g")/("18 g mol"^(-1))=4.44`
`"Mole fraction of KI"=("No. of moles of KI")/("Total no. of moles in solution")=(0.120)/(0.120+4.44)=(0.120)/(4.560)=0.0262`.