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)`.

Explanation

(a) Molar mass of KI `= 39+127=16 g mol^(-1)`
20% (mass/mass) aqueous solution of KI means 20 kg of KI is present in 100 g of solution.
That is,
20 g of KI is present in `(100 - 20)` g of water
= 80 g of water
Therefore, molality of the solution
`= ("Moles of KI")/("Mass of water in kg")=((20)/(166))/(0.08)=1.506 m`
= 1.51 m (approximately).
(b) It is given that the density of hte solution
`= 1.202 g mL^(-1)`
`therefore` Volume of 100 g solution
`= ("Mass")/("Density")=(100 g)/(1.202 g mL^(-1))`
= 83.19 mL
`= 83.19xx10^(-3)L`
Therefore, molarity of the solution
`= ((20)/(160))/(83.19xx10^(-3)L)=1.45M`
(c ) Moles of KI `= (20)/(166)=0.12 mol`
Moles of water `= (80)/(18)=4.44 mol`
Mole fraction of KI `= (0.12)/(0.12+4.44)=0.0263`.