A solution containing 15 g of urea (Molar mass =60 g `mol^(-1)`) per litrea has the same osmotic presure (isotonic) as a solution of glucose (molar mass =180 g `mol^(-1)`) in water, Calculate themass of glucose present in one litre of the solution.

Answer

45 g

Explanation

`"Osmotic pressure"(pi)=(W_(B)xxRxxT)/(M_(B)xxV)`
`"For urea solution "(pi_(1))=((15g)xxRxxT)/((60" g mol"^(-1))xx(1L))`
`"For glucose solution"(pi_(2))=(W_(B)RxxT)/((180" g mol"^(-1))xx(1L))`
`"For isotinic solutions", pi_(1)=pi_(2)`
`therefore((15g))/((60" g mol"^(-1)))=W_(B)/((180" g mol"^(-1)))or BW_(B)=((15g)xx(180" g mol"^(-1)))/((6.0" g mol"^(-1)))=45 g`
`therefore "Mass of glucose"(W_(B))=45.0 g.`