Verified Answer :

moles CO = 10 g x 1mol 28g = 0.357 moles in 3 L = 0.119 mol/L at STP 
P1/n1 = P2/n2 
101.325 kPa/0.119 = 200 kPa/n2 
n2 = 0.235 moles/L = 0.235mol x 28g/mol = 6.6g/L (assuming no change in temp) 

Answer : The solubility of carbon monoxide gas at pressure 200 kPa is, 6.579 g/L

Explanation :

According top the Henry's Law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.

\(S\propto P\)

or,

\(\frac{S_1}{S_2}=\frac{P_1}{P_2}\)

where,

\(S_1\) = initial solubility of gas = \(\frac{10g}{3L}\)

\(S_2\) = final solubility of gas = ?

\(P_1\) = initial pressure of gas = 101.325 kPa (at STP)

\(P_2\) = final pressure of gas = 200 kPa

Now put all the given values in the above formula, we get the final solubility of the gas.

\(\frac{\frac{10g}{3L}}{S_2}=\frac{101.325kPa}{200kPa}\)

\(S_2=6.579g/L\)

Therefore, the solubility of carbon monoxide gas at pressure 200 kPa is, 6.579 g/L

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