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 mixing, total volume of the solution `=100+100=200cm^(3)`
Osmotic pressure due to the urea in the solution :
1.5 g urea which was present originally in `100 cm^(3)` is now present in `200cm^(3)`, i.e., in the final solution
Concentration of urea, `C = 1.5g//200cm^(3)=1.5xx"5 g/litre = 7.5 g/litre "=(7.5)/(60)"moles/litre"`
`" "(because" Molar mass of urea = 60 g mol"^(-1))`
`T = 20+273=293K, R = "0.082 litre atm/degree/mole"`
`therefore" "pi=CRT=((7.5)/(60)" mol L"^(-1))=(0.082" L atm K"^(-1)"mol"^(-1))xx(293 K)=3.00 atm.`
Osmotic pressure due to cane-sugar in the solution : In the final solution.
Concentration of cane sugar, `C=3.42g//200cm^(3)=3.42xx"5 g/litre = 17.10 g/litre"=(17.10)/(342)" moles/litre"`
`" "(because" Molar mass of sugar = 342 g mol"^(-1))`
`therefore" "pi=CRT=((17.10)/(342)" mol L"^(-1))xx("0.082 L atm K"^(-1)"mol"^(-1))(293 K)="1.20 atm"`
As the two solutes behave independent of each other in the solution,
Total Osmotic pressure = Osmotic pressure of urea `+` Osmotic pressure of sugar `=3.00 +1.20=4.20` atm.
Alternatively, as colligative properties depend only on the number of moles of the solute and do not depend upon the nature of the solute, problem can be solved by finding the total number of the moles of the solute as follow :
No. of moles of urea present `=(1.5)/(60)=0.025" , No. of moles of cane sugar present"=(3.42)/(342)=0.01`
Total no. of moles of the solute `=0.025+0.01=0.035`
Total volume of the solution `=100+100cm^(3)=200cm^(3)=0.2L`
Applying the relation, `pi=(n)/(V)RT,` we have `pi=("0.035 mol"xx"0.0821 L atm K"^(-1)"mol"^(-1)xx293K)/(0.2L)="4.20 atm."`