Calculate the osmotic pressure of a solution obtained by mixing 100 mL of 1.6 percent solution of cane sugar (molecular mass = 342 at 293 K) with 3.4% solution of urea (molecular mass = 60).

Answer

7.37. atmosphere

Explanation

Molecular mass of cane sugar = 342, T = 293 K ,
R = 0.0821 `"L atm mol"^(-1)K^(-1)`
Molecular mass of urea = 60,
Total volume of solution `=100+100=200` mL
Molar concentration of cane sugar
`=(1.6)/(200)xx(1000)/(342)"mol L"^(-1)=(8)/(342)M`
Molar concentration of urea `=(3.4)/(200) xx (1000)/(60)=(17)/(60)M`
`pi_(1)` (osmotic pressure of cane sugar)
`=[(8)/(342) xx 0.0821 xx 293]"atm" =0.56` atm
`pi_(2) `(osmotic pressure of urea)
`=[(17)/(60) xx 0.0821 xx 293]"atm"=6.81` atm
Total osmotic pressure `=pi_(1)+pi_(2)=0.56+6.81`
= 7.37 atm.