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

Step 1. Calculation of molality of solution
Weight of `KI` in 100 g of solution = 20 g
Weight of water in solution `= 100 - 20 = 80 kg = 0.80 kg`
Molar mass of `KI = 39 + 127 = 166 "g mol"^(-1)`
Molality of solution `(m)=("No. of gram moles of KI")/("Mass of water in kg")=((20g)//(166"g mol"^(-1)))/((0.08 kg))`
`=1.506 "mol kg"^(-1)=1.506m`
Step II. Calculation of molarity of solution
Weight of solution = 100 g , Density of solution `= 1.202 "g mL"^(-1)`
Volume of solution `= ("Weight of solution")/("Density")=((100g))/((1.202"g mL"^(-1)))=83.19 "mL"`
Molarity of solution `(M)=("No. of gram moled of KI")/("Volume of solution inlitres")=((20g)//(166"g mol"^(-1)))/((0.083L))`
`=1.45 "mol L"^(-1)=1.45M`
Step III. Calculation of mole fraction of `KI`
`n_(KI)=("Mass of KI")/("Molar mass of KI")=((20g))/((166" g mol"^(-1)))=0.12` mol
`n_(H_(2)O)=("Mass of water")/("Molar mass of water")=((80 g))/((18 "g mol"^(-1)))=4.44` mol
`x_(KI)=(n_(KI))/(n_(KI)+n_(H_(2)O))=((0.12 "mol"))/((0.12+4.44)"mol")=(0.12)/(4.56)=0.0263`.