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Question: How many millimoles of \[{{N}_{2}}\] gas would dissolve in\[1\] litre of water if \[{{N}_{2}}\] gas ...

How many millimoles of N2{{N}_{2}} gas would dissolve in11 litre of water if N2{{N}_{2}} gas is bubbled through water at 293K293K, and exerts a partial pressure of 0.9870.987 bar. Given that Henry’s law constant for N2{{N}_{2}} at 293K293K, is 76.4876.48kbar.

Explanation

Solution

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)(x).

Complete answer:
The mathematical equation of Henry’s law for N2H2O{{N}_{2}}-{{H}_{2}}O mixture is given as
pN2=KHxN2(1){{p}_{N2}}={{K}_{H}}{{x}_{N2}}-----(1)
Where pN2{{p}_{N2}}describes the partial pressure of nitrogen gas, KH{{K}_{H}} is the Henry’s constant and xN2{{x}_{N2}}represents mole fraction of N2{{N}_{2}} in N2H2O{{N}_{2}}-{{H}_{2}}Omixture.
Mole fraction describes the ratio of the number of moles of one component in the mixture. Therefore, mole fraction of N2{{N}_{2}} in N2H2O{{N}_{2}}-{{H}_{2}}O mixture is given as
xN2=No.ofmolesofN2No.ofmolesofN2+No.ofmolesofH2O{{x}_{N2}}=\dfrac{No.ofmolesofN{}_{2}}{No.of moles of{{N}_{2}}+No.of moles of{{H}_{2}}O}
As No.ofmolesofH2O=100018=55.55No.of moles of{{H}_{2}}O=\dfrac{1000}{18}=55.55
Put theses value in the equation(1)(1), we get
pN2=KHNo.ofmolesofN2No.ofmolesofN2+55.55{{p}_{N2}}={{K}_{H}}\dfrac{No.ofmolesofN{}_{2}}{No.of moles of{{N}_{2}}+55.55}
The given values are KH=76.48kbar=76480bar{{K}_{H}}=76.48kbar=76480bar and pN2=0.987bar{{p}_{N2}}=0.987bar
Put this value in the above equation, we get
0.987=(76480)No.ofmolesofN2No.ofmolesofN2+55.550.987=(76480)\dfrac{No.ofmolesofN{}_{2}}{No.of moles of{{N}_{2}}+55.55}
By solving the above equation, we get
No.ofmolesofN2=0.716mmolNo.ofmolesofN{}_{2}=0.716mmol
Therefore, 0.716mmol0.716mmolof N2{{N}_{2}}gas would dissolve in 1 litre of water if N2{{N}_{2}} gas is bubbled through water at 293K293K, and exerts a partial pressure of 0.9870.987bar.

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. 0.716mmol0.716mmolof N2{{N}_{2}}gas would dissolve in 11 litre of water if N2{{N}_{2}} gas is bubbled through water at 293K293K, and exerts a partial pressure of 0.9870.987bar. 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.