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1 answer:

4 0

Answer:

umm...what is the question though?

Explanation:

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ivolga24 [154]

Answer : The solubility is, $1.4\times 10^{-5}mol/L$

Explanation : Given,

$k_H$ = Henry's law constant of argon = $1.5\times 10^{-3}mol/L.atm$

First we have to calculate the pressure of argon.

Pressure of argon = $1.0atm\times \frac{0.93}{100}$

Pressure of argon = 0.0093 atm

Now we have to calculate the solubility.

As, the solubility of argon in 1 atm pressure = $1.5\times 10^{-3}mol/L$

So, the solubility of argon in 0.0093 atm pressure = $\frac{0.0093atm}{1atm}\times 1.5\times 10^{-3}mol/L$

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Thus, the solubility is, $1.4\times 10^{-5}mol/L$

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