1.

18 g glucose, C_(6)H_(12)O_(6), is dissolved in 1 kg of water in saucepan. At what temperasture will water boil at 1.013 bar ? K_(b) for water is 0.52 K kg mol^(-1).

Answer»

Solution :Moles of GLUCOSE `= 18g//180 g mol^(-1)=0.1` mol
Number of kilograms of SOLVENT = 1 kg
Thus MOLALITY of glucose solution `= 0.1 mol kg^(-1)`
For water, change in BOILING point
`Delta T_(b)=K_(b)xx m`
`=0.52 "K kg mol"^(-1)xx 0.1 mol kg^(-1) = 0.052` K
Since water boils at 373.15 K at 1.013bar pressure.Therefore, the boiling point of solution will be `373.15+0.052=373.202 K`.


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