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18 g of glucose (C(6)H(12)O(6)) is disso...

`18 g` of glucose `(C_(6)H_(12)O_(6))` is dissolved in `1 kg` of water in a saucepan. At what temperature will the water boil (at 1 atm) ? `K_(b)` for water is `0.52 K kg mol^(-1)`.

Text Solution

Verified by Experts

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`
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