#### Question

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K_{2}SO_{4} in 2 liter of water at 25° C, assuming that it is completely dissociated.

#### Solution

When K_{2}SO_{4} is dissolved in water, K+ and `SO_4^(2-)` ions are produced.

`K_2SO_4` -->`2K^(+) + SO_4^(2-)`

Total number of ions produced = 3

*∴ i *=3

Given,

*w* = 25 mg = 0.025 g

*V* = 2 L

*T* = 25^{0}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}

Appling the following relation,

`pi = i n/vRT`

` iw/M1/v RT`

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

`= 5.27 xx 10^(-3) atm`

Is there an error in this question or solution?

Solution 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. Concept: Colligative Properties and Determination of Molar Mass - Osmosis and Osmotic Pressure.