Determine the osmotic pressure of a solution prepared by dissolving 25 mg of `K_2SO_4` in 2 litre of water at `25^@C` , assuming that it is completely dissociated.

Explanation

` K_2SO_4` dissolved = 25 mg = 0.025 g `T = 25^@C = 298K`
Molar mass of `K_2SO_4 = 2 xx 39 + 32 + 4 xx 16 = 78 + 32 + 64 = 174 g "mol"^(-1)`
As `K_2SO_4` dissociates completely as `K_2SO_4 to 2K^(+) + SO_4^(2-)`
i.e., ions produced = 3 ` therefore i=3`
Applying the following equation
`pi = iCRT = i n/V RT =i xx w/W xx 1/V RT`
Substituting the values, we get
`pi = 3 xx (0.025 g)/(174 g "mol"^(-1)) xx (1)/(2L)xx 0.0821 L atm K^(-1) "mol"^(-1) xx 298 K`
`= 5.27 xx 10^(-3) atm`