18 g glucose and 6 g urea are dissolved in 1 litre aqueous solution at 27∘C. The osmotic pressure of the solution will be:
We know that colligative properties depend only on the number of particles of solute and not on the type or nature of the solute present. Therefore, the amount of solute present is our only concern, whether it is glucose or urea.
So, we know
π∗V=n∗R∗T
Let's use n1 and n2 here, to denote moles of glucose and moles of urea.
n=n1+n2
π∗V=(n1+n2)∗R∗T
n1=18180=0.1mol
n2=660=0.1mol
Therefore,π∗(1)=(0.1+0.1)∗0.082∗(273+27)
⇒π=4.296 atm
Rounding off gives 4.3