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

Total volume of a solution `= 100+100=200` mL
`:.` Conc. of urea, `C_(1)=(1.5)/(200) xx 1000`
`=7.5 gL^(-1)=(7.5)/(60)" mol L"^(-1)`
Conc. of cane sugar,
`C_(2)=(3.42)/(200)xx1000=17.1" gL"^(-1)=(17.1)/(342)" mol L"^(-1)`
Osmotic pressure due to urea, `pi_(1)=C_(1)RT=(7.5)/(60)xx 0.082 xx 293 = 3.00` atm.
Osmotic pressure due to cane sugar,
`pi_(2)=C_(2)RT=(17.1)/(342)xx0.082 xx 293 = 1.20` atm.
Total osmotic pressure,
`pi=pi_(1)+pi_(2)=3.00+1.20=4.20` atm