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

Exercise | Q 15 | Page 46

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.

Solution:

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 = ${\displaystyle \frac{△{\text{T}}_{\text{f}}}{{(△{\text{T}}_{\text{f}})}_0}}$

3. π = MRT = ${\displaystyle \frac{{\text{n}}_2\text{RT}}{\text{V}}}$

4. i = ${\displaystyle \frac{\pi }{{\pi }_0}}$

Calculation:

${\displaystyle △{\text{T}}_{\text{f}}={\text{T}}_{\text{f}}^0-{\text{T}}_{\text{f}}}$

${\displaystyle =0^{\circ }\text{C}-(-{0.510}^{\circ }\text{C})}$ = 0.510 °C = 0.510 K

m = 0.15 m = 0.15 mol kg^–1 

Now, using formula (i),

Δ T_f = K_fm

${\displaystyle {(△{\text{T}}_{\text{f}})}_0}$ = 1.86 K kg mol^-1 × 0.15 mol kg^-1 = 0.279 K

Now, using formula (ii),

i = ${\displaystyle \frac{△{\text{T}}_{\text{f}}}{{(△{\text{T}}_{\text{f}})}_0}=\frac{0.510\text{K}}{0.279\text{K}}}$ = 1.83

Now, using formula (iii),

(π)₀ = MRT

${\displaystyle =\frac{{\text{n}}_2}{\text{V}}\text{RT}}$

${\displaystyle =\frac{0.15 \text{mol}\times 0.08205 {\text{dm}}^3 \text{atm}\cdot {\text{mol}}^{-1}{\text{K}}^{-1}\times 273\text{K}}{1{\text{dm}}^3}}$

= 3.36 atm

Now, using formula (iv),

i = ${\displaystyle \frac{\pi }{{\pi }_0}}$ = 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.