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

Calculatetion of molality of solution
Weight of water in the solution= 20 g
Weight of water in the solution=100-20=80g=0.08 kg
Molar mass of KL=39+127=166g `mol^(-1)`.
`"Molality of solution (m)"=("No of gram moles of KL")/("Mass of water in kg")=((20g)//(166g mol^(-1)))/((0.08 kg))`
Calucalation of molarity of solution
` "Weight of solution"=100g , "Density of solution" = 1.202g mL^(-1)`.
`"Volume of solution"=("Weight of solution")/("Density")=((100g))/((1.202g mL^(-1)))=83.19 mL=0.083L`
`"Molarity of solution (m)"=("No. of gram moles of KL")/("Voluume of solution in litres")=((20g)//(166g mol^(-1)))/((0.083L))`
Calculation of mole fraction of KL
`n_(KL)=("Mass of KL")/("Molar mass of KL")=((20g))/((166g mol^(-1)))=0.12 mol`
`n_(H_(2)O)=("Mass of water")/("Molar mass of water")=((80g))/((18g mol^(-1)))=4.44 mol`
`x_(KL)=n_(KL)/(N_(KL)+n_(H_2O))=((0.12 mol))/((0.12+4.44)mol)=(0.12)/(4.56)=0.0263.`