(a) Explain why on addition of 1 mole glucose to 1 litre water the boling point of water increases .5 (b) Henry's law constant for CO_(2) in water is 1.67cc10^(8)Pa at 298K.Calculate the number of moles of CO_(2) in 500 ml of soda water when packed under 2.53xx10^(5) Pa at same temperature.

(a) Explain why on addition of 1 mole glucose to 1 litre water the bol

1.	(a) Explain why on addition of 1 mole glucose to 1 litre water the boling point of water increases .5 (b) Henry's law constant for CO_(2) in water is 1.67cc10^(8)Pa at 298K.Calculate the number of moles of CO_(2) in 500 ml of soda water when packed under 2.53xx10^(5) Pa at same temperature.
Answer» <P> Solution :(a) Boiling point of solution having 1 mol of glucose is higher than that of pure solvent because the VAPOUR pressure of solution is lower than that of pure solvent and vapour pressure increases with increase in TEMPERATURE. Hence, the solution has to be heated more to make the vapour pressure equal to atmospheric pressure. (b) `K_(H)=1.67xx10^(8)Pa` Henry's law, `P_(CO_(2))=2.53xx10^(5)Pa` `P_(CO_(2))=K_(H)xxx_(CO_(2))` `x_(CO_(2))=(P_(CO2))/(K_(H))=(2.53xx10^(5))/(1.67xx10^(8))` `n_(CO2)/(n_(H2O))=1.525xx10^(-3)` `nH_(2)O=(500)/(18)=27.78[500ml~~500g]` `n_(CO2)=1.515xx10^(-3)xx27.78` `n_(CO2)=42.01xx10^(-3)"moles"`