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Example · Example 2

Q.The partial pressure of CO2\text{CO}_2 gas maintained above a bottle of soft drink at 298 K298\ \text{K} is 4 bar4\ \text{bar}. If the Henry's law constant for CO2\text{CO}_2 dissolved in water at 298 K298\ \text{K} is KH=1.67×103 barK_H = 1.67 \times 10^3\ \text{bar}, calculate the mole fraction of CO2\text{CO}_2 in the drink.

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By Henry's law, p=KHxp = K_H x, so x=p/KHx = p/K_H. Substituting the given values, x=4 bar/(1.67×103 bar)=4/1670≈2.395×10−3x = 4\ \text{bar} / (1.67 \times 10^3\ \text{bar}) = 4/1670 \approx 2.395 \times 10^{-3}. [!ANSWER] The mole fraction of dissolved CO2\text{CO}_2 in the drink is x≈2.40×10−3x \approx 2.40 \times 10^{-3}.

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