A solution of glycerol `(C_(3) H_(8) O_(3))` in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of `100.42^(@)` C while pure water boils at `100^(@)`C. What mass of glycerol was dissolved to make the solution ? `(K_(b) "for water " = 0.512 K "kg mol"^(-1))`

Explanation

`(W_(B))` = Mass of glycerol = ?
`(W_(A))` = Mass of solvent = 500 g
Solution was prepared by dissolving some glycerol in water.
(100.42-100)
`Delta T_(b) = 0.42^(@)`C
`K_(b) = 0.51 " kg mol"^(-1)`
`Delta T_(b) = K_(b) xx `Molality.
Molality (m) = `(W_(b) xx 1000)/(W_(A) xx M_(B))`
Molecular wt. of Glycerol `(M_(B)) C_(3) H_(8) O_(3)`
= ` (12 xx 3) + (1 xx 8) + (16 xx 3)`
36 + 8 + 48 = 92
`Delta T_(b) = (W_(B) xx 1000)/(500 xx 92 ) xx K_(b)`
`0.42 = (0.512 xx W_(B) xx 1000)/(500 xx 92 )`
`W_(B) = (0.42 xx 500 xx 92 )/(0.512 xx 1000)`
`W_(B) = (19320)/(512)`
`W_(B) `= 37.88 g
Weight of glycerol = 37.88 g