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

Answer

91.8

Explanation

`pi = (nRT)/(V)`
`pi ("glycerine") = (10.2)/(M) xx (RT)/(1)`
`pi ("glucose") = (2)/(180) xx (RT)/(0.1)`
`(10.2)/(M) xx (RT)/(1) = (2)/(180) xx (RT)/(0.1)`
` M = (10.2 xx 180 xx 0.1)/(2)`
`= 91.8 g mol^(-1)`