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Henry's law constant for CO_2 in water is 1.67 xx 10^8 Pa at 298 K. Calculate the quantity of CO_2 in 500 mL of soda water when packed under 2.5 atm CO_2pressure at 298 K.

Answer»

Solution :`K_H = 1.67 xx 10^8 Pa, P_(CO_2)` = 2.5 ATM = `2.5 xx 101325 Pa`
APPLYING Henry.s law,
`P_(CO_2) = K_H xx x_(CO_2) " or " x_(CO_2) = (P_(CO_2))/(K_H)`
Substituting the values, we get
` x_(CO_2) = (2.5 xx 101325 Pa)/(1.67 xx 10^8 Pa) = 1.517 xx 10^(-3)`
`(n_(CO_2))/(n_(H_2O) + n_(CO_2)) = (n_(CO_2))/(n_(H_2O))= 1.517 xx 10^(-3)`
For 500 mL of soda water, water present
` = 500/18 = 27.78 `moles i.e., `n_(H_2O) = 27.78` moles
` therefore (n_(CO_2))/(27.78) = 1.517 xx 10^(-3)`
`n_(CO_2)= 27.78 xx 1.517 xx 10^(-3) `mole = `42.14 xx 10^(-3) xx 44 g = 1.854 g `


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