Calculte the osmotic pressure os a solution obtained by mixing 100 mL of 4.5% solution of urea (mol. Mass =60) and 100 mL pf 3.42% solution of cane sugar (mol. Mass =342) at 300 K. `(Given R=0.0821 L atm K^(-1) mol^(-1))`

Explanation

Calculaion of the osmotic pressure of solution containing urea.
`pi_(1)=(W_(B)xxRxxT)/(M_(B)xxV)`
`W_(B)=4.5g,M_(B)=60g mol^(-1),T=300 K`
`R=0.0821 L atm K^(-1)mol^(-1),V=(100+100)=((200mL))/1000=0.2L`
`pi_(1)=((4.5)xx(0.0821 Latm K^(-1)mol^(-1))xx(300K))/((60 g mol^(-1))xx(0.2L))=9.236 atm`
Calculation of the osmotic pressure os solution containing sucrose.
`W_(B)=3.42g, M_(B)=342g mol^(-1), T=300K`
`R=0.0821 L atm K^(-1), V=(100+100)=((200mL))/1000=0.2L`
`pi_(2)=((3.42g)xx(0.0821 L atm K^(-1)mol^(-1))xx(300K))/((342 g mol^(-1))xx(0.2L))=1.23 atm`
Calculation of the total osmotif pressure of soutionn.
Total osmotic pressure `(pi)=pi_(1)+pi_(2)=(9.236+1.23)=10.47 `atm.
No. of moles os urea in the solution`=((4.5g))/((60.0 g mol^(-1)))=0.075` mol
No. of moles of sugar in the solution`=((3.42g))/((342.0 g mol^(-1)))=0.01 mol`
Total volume of the solution=(100+100)=200mL= 0.2 L
Osmotic pressure os the resulting solution `(pi)=CRT=pi/VRT`
`=((0.085 mol)xx(0.0821 L atmK^(-1)mol^(-1))xx(300K))/((0.2L))=10.47 atm`