Question

A 0.15 m aqueous solution of KCl freezes at - 0.510 °C. Calculate i and osmotic pressure at 0 °C. Assume the volume of solution equal to that of water.

Answer 1

Given:
Molality of solution = m = 0.15 m
Freezing point of solution = T_f = - 0.510 °C
Temperature = 0 °C = 273 K

To find:
1. The value of van’t Hoff factor (i)
2. Osmotic pressure of solution

Formulae:
1. Δ T_f = K_fm

2. i = `(triangle "T"_"f")/(triangle "T"_"f")_0`

3. π = MRT = `("n"_2 "RT")/"V"`

4. i = `pi/pi_0`

Calculation:

`triangle "T"_"f" = "T"_"f"^0 - "T"_"f"`

`= 0^circ "C" - (- 0.510 ^circ "C")` = 0.510 °C = 0.510 K

m = 0.15 m = 0.15 mol kg^–1 

Now, using formula (i),

Δ T_f = K_fm

`(triangle "T"_"f")_0` = 1.86 K kg mol^-1 × 0.15 mol kg^-1 = 0.279 K

Now, using formula (ii),

i = `(triangle "T"_"f")/(triangle "T"_"f")_0 = (0.510 "K")/(0.279 "K")` = 1.83

Now, using formula (iii),

(π)₀ = MRT

`= "n"_2/"V" "RT"`

`= (0.15  "mol" xx 0.08205  "dm"^3  "atm" * "mol"^-1 "K"^-1 xx 273 "K")/(1 "dm"^3)`

= 3.36 atm

Now, using formula (iv),

i = `pi/pi_0` = 1.83

π = 1.83 × 3.36 atm

π = 6.15 atm

∴ The van’t Hoff factor is 1.83.

∴ The osmotic pressure of solution at 0 °C is 6.15 atm.