The concentration of \[C{{O}_{2}}\] in a soft drink bottled with a partial pressure of \[C{{O}_{2}}\] of \[4.0\] atm over the liquid at \[25{}^\circ C\] is \[1.2\times {{10}^{-1}}M\]. The Henry's law constant for \[C{{O}_{2}}\] at this temperature is ____________.
A.$3.0\times {{10}^{-2}}mol/L-atm$
B.$4.5\times {{10}^{-3}}mol/L-atm$
C.$5.6\times {{10}^{-3}}mol/L-atm$
D.$2.3\times {{10}^{-2}}mol/L-atm$
Answer
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Hint: We know that the Henry law constant is now dependent only on the mole fraction of gas. If gas which has high solubility would have a higher mole fraction of gas in the liquid and one which has low solubility will have a low value of mole fraction of gas dissolved in liquid. Here we know that Henry's constant and mole fraction are inversely related.
Complete answer:
Henry law is stated in terms of pressure and mole fraction. According to which the pressure of the gas over a solution in which gas is dissolved is proportional to the mole fraction of gas dissolved in the solution. According to Henry's law, \[S=KP\]
Where, \[S\] is the solubility in moles per litre; \[K\] is the henry's law constant and \[P\] is the pressure in atm. Now, Substitute values in the above expression we get;
\[1.2\times {{10}^{-1}}={{K}_{H}}\times 4~atm\]
On further solving we get the value of ${{K}_{H}}$;
\[{{K}_{H}}=3.0\times {{10}^{-2}}mol/L-atm\]
Therefore, the correct answer is option A.
Additional Information:
Thus gas which has a higher value of has lower solubility and which has a low value of \[{{K}_{H}}\] having high solubility in liquid. Therefore we can rearrange the given henry law constant according to the increasing solubility order.
Henry law is observed in various biological processes. Respiration and oxygenation of the blood are one of the most common examples of Henry's law. Inhalation of air increases the partial pressure of oxygen in the lung and thus oxygen gas dissolved in the blood which is then further circulated in the body.in the lungs partial pressure of oxygen is high and thus it is less dissolved in lung thus flow n the body, however, the partial pressure of carbon dioxide is low thus it readily dissolves in lungs and exhaled from the body. Henry law is also widely applied in pressure-filled carbonate drinks.
Note:
Remember that be careful with units. You have to convert them carefully so as to avoid any mistakes in the final result. Also remember the basic concepts like molarity, molality, normality, mole fraction, equivalent weight so as to solve these types of problems without any obstacle.
Complete answer:
Henry law is stated in terms of pressure and mole fraction. According to which the pressure of the gas over a solution in which gas is dissolved is proportional to the mole fraction of gas dissolved in the solution. According to Henry's law, \[S=KP\]
Where, \[S\] is the solubility in moles per litre; \[K\] is the henry's law constant and \[P\] is the pressure in atm. Now, Substitute values in the above expression we get;
\[1.2\times {{10}^{-1}}={{K}_{H}}\times 4~atm\]
On further solving we get the value of ${{K}_{H}}$;
\[{{K}_{H}}=3.0\times {{10}^{-2}}mol/L-atm\]
Therefore, the correct answer is option A.
Additional Information:
Thus gas which has a higher value of has lower solubility and which has a low value of \[{{K}_{H}}\] having high solubility in liquid. Therefore we can rearrange the given henry law constant according to the increasing solubility order.
Henry law is observed in various biological processes. Respiration and oxygenation of the blood are one of the most common examples of Henry's law. Inhalation of air increases the partial pressure of oxygen in the lung and thus oxygen gas dissolved in the blood which is then further circulated in the body.in the lungs partial pressure of oxygen is high and thus it is less dissolved in lung thus flow n the body, however, the partial pressure of carbon dioxide is low thus it readily dissolves in lungs and exhaled from the body. Henry law is also widely applied in pressure-filled carbonate drinks.
Note:
Remember that be careful with units. You have to convert them carefully so as to avoid any mistakes in the final result. Also remember the basic concepts like molarity, molality, normality, mole fraction, equivalent weight so as to solve these types of problems without any obstacle.
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