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InterviewSolution
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`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)`. |
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Answer» Moles of glucose `=18g//180"g "mol^(-1)=0.1` mol Number of kilograms of solvent `=1kg` Thus molality of glucose solution `=0.1` mol `kg^(-1)` For water, change in boiling point. `DeltaT_(b)=K_(b)xxm=0.52"K kg "mol^(-1)xx0.1" mol "kg^(-1)=0.052K` Since water oils at 373.15K at 1.013 bar pressure, therefore, the boiling point of solution will be `373.15+0.052=373.202K` |
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