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प्रश्न
Henry’s law constant for CO2 in water is 1.67 × 108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.
Henry’s law constant for CO2 in water is 1.67 × 108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.53 × 105 Pa at the same temperature.
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उत्तर
Given: KH = 1.67 × 108 Pa
`P_(CO_2)` = 2.5 atm = 2.53 × 105 Pa
According to Henry’s law,
`P_(CO_2) = K_H xx x_(CO_2)`
∴ `x_(CO_2) = P_(CO_2)/K_H`
= `(2.53 xx 10^5 Pa)/(1.67 xx 10^8 Pa)`
= 1.515 × 10−3
We can write, `x_(CO_2) = (n_(CO_2))/(n_(CO_2) + n_(H_2O)) = (n_(CO_2))/(n_(H_2O))`
= 1.515 × 10−3 ...[Since, `n_(CO_2)` is negligible as compared to `n_(H_2O)`.]
For 500 mL of soda water, the volume of water = 500 mL; the mass of water = 500 g,
We can write:
= `500/18 "mol of water"`
= 27.78 mol of water
i.e., `n_(H_2O)` = 27.78
∴ `(n_(CO_2))/n_(H_2O) = x_(CO_2)`
`(n_(CO_2))/(27.78) = 1.515 xx 10^(−3)`
= 1.515 × 10−3 × 27.78
`n_(CO_2)` = 42.08 × 10−3 mol
= 0.04214 mol
∴ Mass of CO2 = `n_(CO_2) xx "Molar mass"`
= 0.04214 × 44 g
= 1.854 g
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