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Solutions question

2009 · Shift 1 · Q5
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  5. /2009 · Shift 1 · Q5

Solutions question

2009 · Shift 1 · Q5

JEE AdvancedChemistrySolutionsMCQ+3 / −1
The Henry's law constant for the solubility of N 2{}_22​ gas in water at 298 K is 1.0 ×\times× 10 5{}^55 atm. The mole fraction of N 2{}_22​ in air is 0.8. The number of moles of N 2{}_22​ from air dissolved in 10 moles of water at 298 K and 5 atm pressure is
  1. A
    4.0 ×\times× 10 −4{}^{-4}−4
  2. B
    4.0 ×\times× 10 −5{}^{-5}−5
  3. C
    5.0 ×\times× 10 −4{}^{-4}−4
  4. D
    4.0 ×\times× 10 −6{}^{-6}−6
View written solutionFree

Correct answer: A

This problem requires the application of Dalton's Law of Partial Pressures and Henry's Law to determine the amount of dissolved nitrogen gas in water.

Step-by-Step Solution:

1. Identify the given information:

  • Henry's law constant for N₂, KH=1.0×105K_H = 1.0 \times 10^5KH​=1.0×105 atm.
  • Temperature, T=298T = 298T=298 K.
  • Total pressure of air, Ptotal=5P_{total} = 5Ptotal​=5 atm.
  • Mole fraction of N₂ in air, yN2=0.8y_{N_2} = 0.8yN2​​=0.8.
  • Moles of water (solvent), nH2O=10n_{H_2O} = 10nH2​O​=10 moles.

2. Calculate the partial pressure of N₂ gas: According to Dalton's Law of Partial Pressures, the partial pressure of a gas in a mixture is the product of its mole fraction and the total pressure. Pgas=ygas×PtotalP_{gas} = y_{gas} \times P_{total}Pgas​=ygas​×Ptotal​ For nitrogen (N₂): PN2=yN2×PtotalP_{N_2} = y_{N_2} \times P_{total}PN2​​=yN2​​×Ptotal​ PN2=0.8×5 atm=4 atmP_{N_2} = 0.8 \times 5 \text{ atm} = 4 \text{ atm}PN2​​=0.8×5 atm=4 atm This is the pressure of the nitrogen gas above the water surface.

3. Apply Henry's Law to find the mole fraction of N₂ in the solution: Henry's Law states that the partial pressure of a gas above a liquid is directly proportional to its mole fraction in the liquid solution. Pgas=KH×xgasP_{gas} = K_H \times x_{gas}Pgas​=KH​×xgas​ Where xgasx_{gas}xgas​ is the mole fraction of the gas dissolved in the solution. Rearranging the formula to solve for xN2x_{N_2}xN2​​: xN2=PN2KHx_{N_2} = \frac{P_{N_2}}{K_H}xN2​​=KH​PN2​​​ xN2=4 atm1.0×105 atm=4.0×10−5x_{N_2} = \frac{4 \text{ atm}}{1.0 \times 10^5 \text{ atm}} = 4.0 \times 10^{-5}xN2​​=1.0×105 atm4 atm​=4.0×10−5 This is the mole fraction of N₂ dissolved in water.

4. Calculate the number of moles of dissolved N₂: The mole fraction of N₂ in the solution is defined as: xN2=nN2nN2+nH2Ox_{N_2} = \frac{n_{N_2}}{n_{N_2} + n_{H_2O}}xN2​​=nN2​​+nH2​O​nN2​​​ Where nN2n_{N_2}nN2​​ is the number of moles of dissolved nitrogen and nH2On_{H_2O}nH2​O​ is the number of moles of water.

Since the solubility of N₂ is very low (as indicated by the small value of xN2x_{N_2}xN2​​), the number of moles of dissolved N₂ (nN2n_{N_2}nN2​​) will be very small compared to the number of moles of water (nH2On_{H_2O}nH2​O​). Therefore, we can make the approximation: nN2+nH2O≈nH2On_{N_2} + n_{H_2O} \approx n_{H_2O}nN2​​+nH2​O​≈nH2​O​ So, the mole fraction expression simplifies to: xN2≈nN2nH2Ox_{N_2} \approx \frac{n_{N_2}}{n_{H_2O}}xN2​​≈nH2​O​nN2​​​ Now, we can solve for nN2n_{N_2}nN2​​: nN2≈xN2×nH2On_{N_2} \approx x_{N_2} \times n_{H_2O}nN2​​≈xN2​​×nH2​O​ nN2≈(4.0×10−5)×(10 moles)n_{N_2} \approx (4.0 \times 10^{-5}) \times (10 \text{ moles})nN2​​≈(4.0×10−5)×(10 moles) nN2≈4.0×10−4 molesn_{N_2} \approx 4.0 \times 10^{-4} \text{ moles}nN2​​≈4.0×10−4 moles

Conclusion: The number of moles of N₂ from air dissolved in 10 moles of water is 4.0×10−44.0 \times 10^{-4}4.0×10−4. This corresponds to option A.

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