Calculate (i) molality (ii) molarity and (iii) mole fraction of KI, if the density of 20% (mass /mass) aq. KI is 1.202 g `mol^(-1)`.

Explanation

(i) Molality (m)
Weight of KI in 100 g of water = 20g
Weight of water in the solution
`= (100 - 20)g = 80g = 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`
(ii) Molarity (M)
Weight of the solution = 100 g
Density of the solution = `1.202 g mL^(-1)`
Volume of the solution `= ("Weight of solution")/("Density")`
`= ((100 g))/((1.202 g mL^(-1))) = 83.19 mL = 0.083 L`
Molarity of the solution (M)
`= ("Number of gram moles of KI")/("Volume of solutions in litres")`
`((20g)//(166 g mol^(-1)))/((0.083 L)) = 1.45 mol L^(-1) = 1.45 M`
(iii) Mole fraction of KI
Number of moles of KI,
`n_(KI) = ("Mass of KI")/("Molar mass of KI") = (20g)/(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
`chi_(KI) = (n_(KI))/(n_(KI) + n_(H_(2)O)) = ((0.12 mol))/((0.12 + 4.44) mol) = (0.12)/(4.56) = 0.0263`