At 25 °C, a 0.1 molal solution of CH3COOH is 1.35 % dissociated in an aqueous solution. Calculate freezing point and osmotic pressure of the solution assuming molality and molarity to be identical. - Chemistry

Exercise | Q 14 | Page 46

At 25 °C, a 0.1 molal solution of CH3COOH is 1.35 % dissociated in an aqueous solution. Calculate the freezing point and osmotic pressure of the solution assuming molality and molarity to be identical.

Solution

Given: 
Molality of solution (m) = 0.1 m = 0.1 mol kg^-1
Degree of dissociation (α) = 1.35% = 0.0135
Temperature = 25 °C = 25 °C + 273.15 = 298.15 K
Molarity of solution (M) = 0.1 M

To find: 
1. Freezing point of solution
2. Osmotic pressure of solution

Formulae:
1. α = ${\displaystyle \frac{\text{i}-1}{\text{n}-1}}$

2. Δ T_f = i K_fm

3. π = i MRT

Calculation:

Using formula (i),

α = ${\displaystyle \frac{\text{i}-1}{\text{n}-1}=\text{i}-1}$ because n = 2

∴ i = 1 + α = 1 + 0.0135 = 1.0135

Now, using formula (ii),

Δ T_f = i K_fm = 1.0135 × 1.86 K kg mol^-1 × 0.1 mol kg^-1 

= 0.189 K = 0.189 °C

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

∴ ${\displaystyle {\text{T}}_{\text{f}}={\text{T}}_{\text{f}}^0-{\text{T}}_{\text{f}}=0^{\circ }\text{C}-\left({0.189}^{\circ }\text{C}\right)}$ = - 0.189 °C

Now, using formula (iii),

π = i MRT = 1.0135 × 0.1 mol dm^-3 × 0.08205 dm³ atm K^-1 mol^-1 × 298.15 K = 2.48 atm

∴ The freezing point of the solution is – 0.189 °C

∴ The osmotic pressure of solution at 25 °C is 2.48 atm