Question

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 liter of water at 25°C, assuming that it is completely dissociated.

Answer 1

When K₂SO₄ is dissolved in water, K⁺ and \[\ce{SO^{2-}4}\] ions are produced.

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

Total number of ions produced = 3

∴ i = 3

Given,

w = 25 mg = 0.025 g

V = 2 L

T = 25°C = (25 + 273) K = 298 K

Also, we know that:

R = 0.0821 L atm K^-1 mol^-1

M = (2 × 39) + (1 × 32) + (4 × 16) = 174 g mol^-1

Applying the following relation,

`pi = "i" "n"/"V""RT"`

= `"i" xx "w"/"M" xx 1/"V""RT"`

= `3xx0.025/174 xx 1/2xx0.0821xx298`

= 5.27 × 10⁻³ atm