At what concentration of solute 'P' the solution will be isotonic with a 5 percent solution of urea? (Molar mas of 'P'=342g `mol^(-1)`)

Explanation

Since solutions are isotonic, `pi_(solute)=pi_(urea)`
`pi_(urea)=(W_(B)xxRxxT)/(M_(B)xxV)`
`W_(B)=5g,M_(B)=(60 g mol^(-1),V=0.1 L`
`pi_(urea)=((5g)xxRxxT)/((60g mol^(-1))xx(0.1L))`
`pi_(solute)=(W_(B)xxRxxT)/(M_(B)xxV)`
`M_(B)=(342 g mol^(-1)),V==0.1 L`
`pi_(solute)=(W_(B)xxRxxT)/((324g mol^(-1))xx(0.1L))`
Equating eqns. (i) and (ii),
`((5g)xxRxxT)/((60g mol^(-1)xx(0.1L)))=(W_(B)xxRxxT)/((342g mol^(-1))xx(0.1 L))`
`W_(B)=((5g)xx(342g mol^(-1)))/((60g mol^(-1)))=28.5 g`
Concentation of solute 'P'=28.5g/100 mL or 28.5%