Calculate the osmotic pressure of a solution obtained by mixing `100 cm^(3)` of `1.5%` solution of urea (mol. Mass=60) and `100 cm^(3)` of `3.42%` solution by cane sugar (mol. Mass = 342) at `20^(@)C`. (R=0.082 litre atm/deg/mole)

Explanation

After mole total volume of the solution `=100 + 100=200 cm^(3)`
Osmotic pressure due to the urea in the solution.
`1.5 g` of urea which was present originally in `100 cm^(3)` is now present in `200 cm^(3)`,i.e., in the final solution.
`Mw_(1)=Mw_("urea")=(W_(2) xx 1000)/(Mw_(2) xx vol. of sol.)=(1.5 xx 1000)/(60 xx 200)=0.125`
`Mw_(2)=Mw_(sugar)=(3.42 xx 1000)/(342 xx 200)=0.05`
`pi=(Mw_(1)+Mw_(2))RT`
`=(0.125 + 0.05) xx0.82 xx 293 =4.2 atm`