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A solution sontaining 10.2 g of elycring pe litre is found to be isotonic with 2% solution of glucose `("molar mass"=180 g mol^(-1))`. Calculate the molar mass of glycrine.

Answer» Since solution are isotonic ,`pi_("glycerine")=pi_("glucose")`
`pi_("glycerine")=(W_(B)xxRxxT)/(M_(B)xxV)`
`W_(B)=10.2g, V=1 L`
`pi_("glycerine")=((10.2g)xxRxxT)/((M_(B))xx(1L))`
`pi_("glucose")=(W_(B)xxRxxT)/(M_(B)xxV)`
` W_(B)=2g, M_(B)=(180 gmol^(-1),V=100 mL=0.1L`
`pi("glucose")=((2g)xxRxxT)/((180g mol^(-1))xx(0.1L))`
Equating eqns, (I) and (II):
` ((10.2g)xxRxxT)/((M_(B))xx(1L))=((2g)xxRxxT)/((180g mol^(-1))xx(0.1L))`
`M_(B)=((10.2g)xx(180g mol^(-1))xx(0.1L))/((1L)xx(2g))=91.8 g mol^(-1)`
Molar mass of glycerisng= 91.8 g `mol^(-1)`


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