18 g glucose, `C_6H_12O_6` (Molar Mass = 180 g `mol^(-1)`) is dessolved in 1 kg of water in a sauce pan. At what temperature will this solution boil ?
`K_b" for water "=0.52 K kg mol^(-1)`, boiling point of pure water =373.15 K)

Explanation

The given values are:
`W_("solute") = 18 g`
`W_("solvent") = 1 kg`
`W_(b) = 0.52 K kg mol^(-1)`
First we calculate elevation in the boiling point of solution.
`DeltaT_(b) = (K_(b) xx W_("solute"))/(Mw_("solute") xx W_("solvent"))`
`= (0.52 xx 18)/(180 xx 1) = 0.052 K`
Since water boils at `373.15 K` at `1 atm` pressure, therefore the boiling point of solution will be
`T_(B) = T_(b)^(@) + DeltaT_(b) = 373.15 + 0.052 = 373.202 K`
Thus , the boiling point of solution is `373.202 K`