A solution of glycerol `(C-(3)H_(8)O_(3)` , molar mass = 92 g `mol^(-1)` iin water was prepared by dissolving some glycero 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

`W_(H_(2)O) = 500 g = 0.5 kg, T_(b)^(0) = 100^(0)CW_("glycerol")`
`:. T_(b) = 100.42 .^(@)C`.
`DeltaT_(b) = 0.42.^(@)C`
`DeltaT_(b) = K_(bm)`
`DeltaT_(b) = (W_("glycerol"))/(W_(H_(2)O) xx M_("glycerol"))`
`0.42 = 0.512 xx (W_("glycerol"))/(0.5 xx 92)`
`W_("glycerol") = (0.42 xx 0.5 xx 92)/(0.512) = 37.73 g`.