Question

The vapour pressure of a solution of urea is 736.2 mm at 100°C. Calculate the osmotic pressure of this solution at 15°C.

Answer 1

Given: Vapour pressure of pure water at 100°C = 760 mm Hg

Vapour pressure of urea solution = 736.2 mm Hg

Temperature for osmotic pressure calculation = 15°C = 288 K

R (gas constant) = 0.0821 L atm mol⁻¹ K⁻¹

Using Raoult’s law

`(Delta P)/P^circ = (P^circ - P)/P^circ = x_"solute"`

⇒ `x_"solute" = (760 - 736.2)/760`

= `23.8/760`

`x_"solute" = 0.0313`

1 mole of water = 18 g = 0.018 kg = 18 mL = 0.018 L

Let number of moles of solute = n

From mole fraction

`x_"solute" = n/(n + 1)`

⇒ `0.0313 = n/(n + 1)`

⇒ 0.0313(n+1) = n

⇒ 0.0313 n + 0.0313 = n

⇒ n(1 − 0.0313) = 0.0313

⇒ `n = 0.0313/0.9687`

n ≈ 0.0323

We know that

`C = n/V`

= `0.0323/0.018`

C = 1.794 mol/L

By using the osmotic pressure formula:

π = CRT

= 1.794 × 0.0821 × 288

= 42.4 atm