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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

Sum

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.

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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 = `("i" - 1)/("n" - 1)`

2. Δ Tf = i Kfm

3. π = i MRT

Calculation:

Using formula (i),

α = `("i" - 1)/("n" - 1) = "i" - 1` because n = 2

∴ i = 1 + α = 1 + 0.0135 = 1.0135

Now, using formula (ii),

Δ Tf = i Kfm = 1.0135 × 1.86 K kg mol-1 × 0.1 mol kg-1 

= 0.189 K = 0.189 °C

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

∴ `"T"_"f" = "T"_"f"^0 - "T"_"f" = 0^circ "C" - (0.189^circ "C")` = - 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

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APPEARS IN

Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 2 Solutions
Exercise | Q 14 | Page 46
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