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2 years ago

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If 120.3 mL of water is shaken with oxygen gas at 2.1 atm, it will dissolve 0.0043 g O2. Estimate the Henry's law constant for t

he oxygen gas in water in units of g mL-1atm-1.

1 answer:

nikklg [1K]2 years ago

7 0

<u>Answer:</u> The Henry's law constant for oxygen gas in water is $1.702\times 10^{-5}g/mL.atm$

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{O_2}=K_H\times p_{O_2}$

where,

$K_H$ = Henry's constant = ?

$C_{O_2}$ = solubility of oxygen gas = $0.0043g/120.3mL$

$p_{O_2$ = partial pressure of oxygen gas = 2.1 atm

Putting values in above equation, we get:

$0.0043g/120.3mL=K_H\times 2.1atm\\\\K_H=\frac{0.0043g}{120.3mL\times 2.1atm}=1.702\times 10^{-5}g/mL.atm$

Hence, the Henry's law constant for oxygen gas in water is $1.702\times 10^{-5}g/mL.atm$

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