18g glucose (C6H12O6) is dissolved in 1kg of water in a saucepan At what temperature will water boil under 1.013 bar pressure? Given kb for water is 0.52kg/mol
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Given
Weight of the glucose is 18g
Mass of the glucose is 180g
Kb for water is 0.52kg/mol
We know that
At 1atm boiling point of water is 100 degree Celsius
We know that
Delta Tb=(Kb*wt.of glucose) /(Wt. Of water * GMW of glucose)
Delta Tb=(0.52*18)/(180*1)
Delta Tb= 9.36/180
Delta Tb=0.052
That impise
Ts-100=0.052
Ts=100.052
Therefore water boiling point is 100.052celsius
Weight of the glucose is 18g
Mass of the glucose is 180g
Kb for water is 0.52kg/mol
We know that
At 1atm boiling point of water is 100 degree Celsius
We know that
Delta Tb=(Kb*wt.of glucose) /(Wt. Of water * GMW of glucose)
Delta Tb=(0.52*18)/(180*1)
Delta Tb= 9.36/180
Delta Tb=0.052
That impise
Ts-100=0.052
Ts=100.052
Therefore water boiling point is 100.052celsius
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