A solution of glycerol `C_(3)H_(8)O_(3)` , molar mass = 92 g `mol^(-1)` in water was prepared by dissolving some glycerol 500 g of water. This solution has a boiling point of `100.42^(@)C`. What mass of glycerol was dissolved to make this solution ? `K_(b)` for water=` 0.512 K kg mol^(-1).`

Explanation

Here `w_(1)=500 g, M_(2)=92 g mol^(-1),w_(2)` = ?
`DeltaT_(b)=(100.42-100) ^(@)C=0.42^(@)Cor0.42 K`
`therefore`" "DeltaT_(b)=K_(b)xx(w_(2)xx1000)/(M_(2)xxw_(1))`
`therefore` " "`0.42=(0.512xxw_(2)xx1000)/(92xx500)`
`Rightarrow`" "`w_(2)=(0.42xx92xx500)/(0.512xx1000)=37.7 g` Therefore, mass of glycerol=37.7 g