Calculate a) molality b) molarity and c) mole fraction of KI if the density of `20%` (mass / mass) aqueous KI is `1.202" g mL"^(-1)`.
As density and % by mass is given, so find the mass of solute and solvent (as x % solution contains x g solute in (100 - x) g solvent).
Find volume of the solution, by using,,
Volume `=("Mass")/("Density")`
Recall the formulae of molality, molarity and mole fraction, to calculate them.
Molality `=("Mass of solute/ molar mass of solute")/("Mass of solventin kg")`

Explanation

a) Molality
Weight of KI in 100 g of water = 20 g
Weight of water in the solution `=100-20=80" g"=0.08" kg"`
Molar mass of KI `=39+127=166" g mol"^(-1)`
Molality of the solution (m) `=("Number of moles of KI")/("Mass of water in kg")`
`=((20"g")//(166" g mol"^(-1)))/((0.08" kg"))=1.506" mol kg"^(-1)=1.506" m"`
b) Molarity ltbbrgt Weight of the solution = 100 g
Denisty of the solution `=1.202" mL"^(-1)`
Volume of the solution `=("Weight of solution")/("Density")=((100" g"))/((1.202" g mL"^(-1)))`
`=83.16" mL"=0.083" L"`
Molarity of the solution (M) `=("Number of gram moles of KI")/("Volume of solution in litres")`
`=((20"g")//(166" g mol"^(-1)))/((0.083" L"))=1.45" mol L-1"=1.45" M"`
c) Mole fraction of KI
(Number of moles of KI)`n_("KI")=("Mass of KI")/("Molar mass of KI")`
`=((20" g"))/((166" g mol"^(-1)))=0.12" mol"`
(Number of moles of water) `n_("H"_(2)"O")=("Mass of water")/("Molar mass of water")`
`=((80" g"))/((18" g mol"^(-1)))=4.44" mol"`
(Mole fraction of KI) `x_("KI")=(n_("KI"))/(n_("KI")+n_("H"_(2)"O"))=((0.12" mol"))/((0.12+4.44)" mol")`
`=(0.12)/(4.56)=0.0263`