Osmotic pressure of a solution obtained by mixing `100 mL` of 1.4% solution of urea (mol mass = 60) and `100 mL` of `3.42%` of cane sugar solution (mol mass = 342) at `20^(@)C (R = 0.0821 L atm K^(-1) mol^(-1))`

Explanation

After mixing the volume of the solution = 200 mL
(i) Osmotic pressure of urea .
`w_(B) =3.4 g , V = 200 mL = 0.2 L , T = 293 K `,
`M_(B) = 60 , R = 0.083 L bar mol^(-1) K^(-1)`
`pi = (w_(B)RT)/(M_(B)V)`
or `" " pi = (3.4 xx 0.083 xx 293)/(60 xx 0.2) = 6.89 bar `
(ii) Osmotic pressure of cane sugar
`w_(B) = 1.6 g V = 200 mL = 0.2 L T = 293 K, M _(B) = 342 ` `R = 0.083 L bar mol^(-1) K^(-1)`
`pi = (w_(B) RT)/(M_(B)V)`
or `" " pi = (1.6 xx 0.083 xx 29)/(342xx0.2) = 0.57 bar `
Since the dilute solution behave like ideal gases, the total osmotic pressure will be equal to sum of the partical osmotic pressure (similar to Dalton.s law of partial pressure of gases).
` therefore " " pi = 6.89 + 0.57 = 7.46` bar .