Osmotic pressure of `40%` (wt.`//`vol.) urea solution is `1.64 atm` and that of `3.42%` (wt.`//`vol.) cane sugar is `2.46 atm`. When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is:

Options

A. 1.64 atm
B. 2.46 atm
C. 2.06 atm
D. 0.82 atm

Answer

C

Explanation

`pi_(mu)=1.66` atm, `pi_(S)=2.46` atm
If equal volumes (V) of both solutions are mixed and original molar concentration of urea solution and sugar solutions are `C_(mu)` and `C_(S)` then
`pi_(1)=(C_(mu))/(2V)RT , pi_(2)=(C_(S))/(2V)RT`
`C_(mu)=n_(1)//V , C_(S)=n_(2)//V`
`C_(1)=(n_(1))/(2V)=(C_(mu))/(2) , C_(2)=(n_(2))/(2V)=(C_(S))/(2)`
`pi_(1)=C_(1)RT and pi_(2)=C_(2)RT`
`pi_(1)=(C_(mu))/(2)RT and pi_(2)=(C_(S))/(2)RT`
`pi=pi_(1)+pi_(2)=(C_(muRT+C_(S)RT)//2`
`=(pi_(mu)+pi_(s))/(2)=(1.66+2.46)/(2)=2.06` atm.