Question

How many grams of KCl should be added to 1.00 kilogram of water to lower its freezing point to −8.0°C? K_f(H₂O) = l .88°C kg mo1⁻¹ (At. masses: K = 39, Cl = 35.5).

Answer 1

KCl dissociates completely in solution as

	\[\ce{KCl \phantom{..}<=> \phantom{.}K+ \phantom{.}+ \phantom{..}Cl−}\]
Initially	1 mole               -              -
At equilibrium	-           1 mole    1 mole

∴ `i = "No. of moles in solution"/"No. of moles added"`

= `(1 + 1)/1`

= 2

Suppose w g of KCl are required to be dissolved in 1.00 kg of water. Therefore, molality of the solution

`m = (w//74.5)/1`

= `w/74.5`

ΔT_f required = 0 − (−8.0)

= 8.0°C

According to the modified equation for depression of freezing point,

ΔT_f = i × K_f × m

`8 = 2 xx 1.88 xx w/74.5`

or `w = (8 xx 74.5)/(2 xx 1.88)`

= 158.5 g

Hence, the required mass of KCl is 158.5 g.