18 g of glucose, `C_(6)H_(12)O_(6)` (molar mass `= 180 g mol^(-1)`) is dissolved 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

Given,
`W_(a)` = weight of `H_(2)O` (solvent) = 1kg
`W_(b)` = weight of `C_(6)H_(12)O_(6)` (glucose , solute) = 18 g
`T_(b)^(2) = 373.15 K`
`K_(b) = 0.52 K kg mol^(-1)`
`M_(b)` = Molar mass of solute (glucose) `= 180 g mol^(-1)`
`Delta T_(b) = (K_(b) xx 1000 xx W_(b))/(M_(b) xx W_(a)) = ((0.52 K kg mol^(-1)) xx 1000 xx 18 g)/((180 g mol^(-1)) xx 1000 g)`
`= (9360)/(180000) = 0.052 K`
As we know that,
`Delta T_(b) = T_(b) - T_(b)^(@)`
`0.052 = T_(b) - 373.15`
`T_(b) = 373.15 + 0.052 = 373.202 K`
`~~ 373.20 K` (approx.)