Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2 K. `(K_(f) " for water = " 1.86 " K kg "mol^(-1))`

Explanation

Since one mole of KCI gives 2 mole particles, the value of
I = 2
`Delta T_(f) = 2 K `
`K_(f) = 1.86 " kg mol"^(-1)`
Applying equation, `T_(f) = iK_(f) `m
`m = (Delta T_(f))/(iK_(f)) = (2)/(2 xx 1.86)`
Therefore, 0.54 mole of KCI should be added to one kg of water.
Molar mass of KCI = 39 + 35.5 = 74.5 g
Amount of KCl = 0.54 `xx 74.5 `g = 40.23 g