Calculate the freezing point of the one molar aqueous solution (density `1.04 g L^(-1)`) of KCl (`k_(f)` for wateer ` = 1.86 kg mol^(-1)` , atomic mass of K = 39 ,Cl = 35.5)

Answer

`-3.852^(@)C`

Explanation

1 molar solution means that 1 mole of KCI is dissolved in 1000 mL of solution.
Mass of solution ` = 1000xx 1.04 = 1040 g`
Mass of KCl = 74.5 g
Mass of water = 1040-74.5 = 965.5 g .
Molality of solution `= (1)/(965.5) xx 1000 = 1.0357` m
Since KCl is strong electrolyte , i = 2
`Delta T_(f) =i K_(f) xx m`
` = 2 xx 1.86 xx 1.0357 = 3.852^(@)`
Freezing point of solution ` = 0 - 3.852 = - 3.852^(@)C`