A 10% aqueous solution of cane sugar (molecular mass 342) is isotonic with 1.754% aqueous solution of urea. Find the molecular mass of urea.

Answer

60 g `mol^(-1)`

Explanation

`pi_("glucose")=(W_(B)xxRxxT)/(M_(B)xxV)=((10g)xxRxxT)/((342" g mol"^(-1))xxRxxT)`
`pi_("urea")=(W_(B)xxRxxT)/(M_(B)xxV)=((1.754g)xxRxxT)/(M_(B)xxRxxT)`
`"Since solutions area isotonic," pi_("glucose")=P_("urea")`
`((10g))/((342" g mol"^(-1)))=((1.754 g))/(MB)or M_(B)=((1.754g)xx(342"g mol"^(-1)))/((10g))`
`=59.98~~60" g mol"^(-1)`.