
How many millimoles of \[{{N}_{2}}\] gas would dissolve in\[1\] litre of water if \[{{N}_{2}}\] gas is bubbled through water at \[293K\], and exerts a partial pressure of \[0.987\] bar. Given that Henry’s law constant for \[{{N}_{2}}\] at \[293K\], is \[76.48\]kbar.
Answer
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Hint: In a mixture of gases, the pressure exerted by each gas is called partial pressure. According to Henry’s law, the amount of gas dissolved in one liter of liquid is directly proportional to the partial pressure of the gas at constant temperature. Concentration of gas in liquid can be represented by mole fraction \[(x)\].
Complete answer:
The mathematical equation of Henry’s law for \[{{N}_{2}}-{{H}_{2}}O\] mixture is given as
\[{{p}_{N2}}={{K}_{H}}{{x}_{N2}}-----(1)\]
Where \[{{p}_{N2}}\]describes the partial pressure of nitrogen gas, \[{{K}_{H}}\] is the Henry’s constant and \[{{x}_{N2}}\]represents mole fraction of \[{{N}_{2}}\] in \[{{N}_{2}}-{{H}_{2}}O\]mixture.
Mole fraction describes the ratio of the number of moles of one component in the mixture. Therefore, mole fraction of \[{{N}_{2}}\] in \[{{N}_{2}}-{{H}_{2}}O\] mixture is given as
\[{{x}_{N2}}=\dfrac{No.ofmolesofN{}_{2}}{No.of moles of{{N}_{2}}+No.of moles of{{H}_{2}}O}\]
As \[No.of moles of{{H}_{2}}O=\dfrac{1000}{18}=55.55\]
Put theses value in the equation\[(1)\], we get
\[{{p}_{N2}}={{K}_{H}}\dfrac{No.ofmolesofN{}_{2}}{No.of moles of{{N}_{2}}+55.55}\]
The given values are \[{{K}_{H}}=76.48kbar=76480bar\] and \[{{p}_{N2}}=0.987bar\]
Put this value in the above equation, we get
\[0.987=(76480)\dfrac{No.ofmolesofN{}_{2}}{No.of moles of{{N}_{2}}+55.55}\]
By solving the above equation, we get
\[No.ofmolesofN{}_{2}=0.716mmol\]
Therefore, \[0.716mmol\]of \[{{N}_{2}}\]gas would dissolve in 1 litre of water if \[{{N}_{2}}\] gas is bubbled through water at \[293K\], and exerts a partial pressure of \[0.987\]bar.
Note:
It is important to note that Henry’s law describes the relation between the amount of gas dissolved in one liter of liquid to the partial pressure of the gas at constant temperature. \[0.716mmol\]of \[{{N}_{2}}\]gas would dissolve in \[1\] litre of water if \[{{N}_{2}}\] gas is bubbled through water at \[293K\], and exerts a partial pressure of \[0.987\]bar. Henry is applicable to dilute solutions only and has many practical applications such as determination of the amount of dissolved oxygen in the blood in order to describe the risk of decompression sickness.
Complete answer:
The mathematical equation of Henry’s law for \[{{N}_{2}}-{{H}_{2}}O\] mixture is given as
\[{{p}_{N2}}={{K}_{H}}{{x}_{N2}}-----(1)\]
Where \[{{p}_{N2}}\]describes the partial pressure of nitrogen gas, \[{{K}_{H}}\] is the Henry’s constant and \[{{x}_{N2}}\]represents mole fraction of \[{{N}_{2}}\] in \[{{N}_{2}}-{{H}_{2}}O\]mixture.
Mole fraction describes the ratio of the number of moles of one component in the mixture. Therefore, mole fraction of \[{{N}_{2}}\] in \[{{N}_{2}}-{{H}_{2}}O\] mixture is given as
\[{{x}_{N2}}=\dfrac{No.ofmolesofN{}_{2}}{No.of moles of{{N}_{2}}+No.of moles of{{H}_{2}}O}\]
As \[No.of moles of{{H}_{2}}O=\dfrac{1000}{18}=55.55\]
Put theses value in the equation\[(1)\], we get
\[{{p}_{N2}}={{K}_{H}}\dfrac{No.ofmolesofN{}_{2}}{No.of moles of{{N}_{2}}+55.55}\]
The given values are \[{{K}_{H}}=76.48kbar=76480bar\] and \[{{p}_{N2}}=0.987bar\]
Put this value in the above equation, we get
\[0.987=(76480)\dfrac{No.ofmolesofN{}_{2}}{No.of moles of{{N}_{2}}+55.55}\]
By solving the above equation, we get
\[No.ofmolesofN{}_{2}=0.716mmol\]
Therefore, \[0.716mmol\]of \[{{N}_{2}}\]gas would dissolve in 1 litre of water if \[{{N}_{2}}\] gas is bubbled through water at \[293K\], and exerts a partial pressure of \[0.987\]bar.
Note:
It is important to note that Henry’s law describes the relation between the amount of gas dissolved in one liter of liquid to the partial pressure of the gas at constant temperature. \[0.716mmol\]of \[{{N}_{2}}\]gas would dissolve in \[1\] litre of water if \[{{N}_{2}}\] gas is bubbled through water at \[293K\], and exerts a partial pressure of \[0.987\]bar. Henry is applicable to dilute solutions only and has many practical applications such as determination of the amount of dissolved oxygen in the blood in order to describe the risk of decompression sickness.
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