1.

Henry's law constant for CO_(2) in water is 1.67xx108 Pa at 298 K. Calculate the quantity of CO_(2) in 500 mL of soda water when packed under 2.5 atm CO_(2) presure at 298 K.

Answer»

Solution :It is given that :
`K_(H)=1.67xx10^(8)Pa`
`p_(CO_(2))=2.5 ATM = 2.5xx1.01325xx10^(5)Pa`
`= 2.533125xx10^(5)Pa`
According to Henry.s law :
`p_(CO_(2))=K_(H).chi`
`chi=(p_(CO_(2)))/(K_(H))=(2.533125xx10^(5))/(1.67xx10^(8))=0.00152`
`chi =(n_(CO_(2)))/(n_(CO_(2))+n_(H_(2)O))=(n_(CO_(2)))/(n_(H_(2)O))`
`nCO_(2)`
is neglected as COMPARED to `nH_(2)O` [since],in 500 mL of soda water, the volume of water = 500 mL [Neglecting the AMOUNT of soda present] We can write :
Mole of water `= (500)/(18)=2.78` MOL
`chi=(n_(CO_(2)))/(n_(H_(2)O))=(n_(CO_(2)))/(27.78)=0.00152`
`n_(CO_(2))=0.042 mol`
Hence, quantity of `CO_(2)` in 500 mL of soda water
`= (0.042xx44)G`
= 1.848 g.


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