Question

Determine the osmotic pressure of a solution prepared by dissolving 2.32 × 10⁻² g of K₂SO₄ in 2L of solution at 25°C assuming that K₂SO₄ is completely dissociated.

(R = 0.082 L atm K⁻¹ mol, Molar mass K₂SO₄ = 174 g mol⁻¹)

Answer 1

Amount of K₂SO₄ dissolved = 2.32 × 10⁻² g

Volume of solution = 2 L

Temperature = 25 + 273 = 298 K

Molar mass of K₂SO₄ = (2 × 39) + 32 + 4 × 16 = 174 g/mol

Since K₂SO₄ dissociates completely as

\[\ce{K2SO4 -> 2K^+ + SO^{2-}4}\]

Total ions produced after dissociation = 3

So, i = 3, π = iCRT

C = `4/"V"`

π = `(3 xx 2.32 xx 10^-2)/(174 xx 2) xx 0.082 xx 298`

= `85.14/(174 xx 100)`

= 0.0048

= 4.8 × 10⁻³ atm