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The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 x 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure is. Plz explain how to solve this question. Ans=4×10^-4.plz solve.?
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The Henry’s law constant for the solubility of N2 gas in water at 298 ...
Henry's Law:
Henry's Law states that the amount of gas dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid. It can be mathematically represented as:

C = k * P

Where:
C = concentration of the dissolved gas in the liquid
k = Henry's law constant
P = partial pressure of the gas above the liquid

Given:
Henry's law constant (k) for N2 gas in water at 298 K = 1.0 x 10^5 atm
Mole fraction of N2 in air = 0.8
Number of moles of water = 10
Pressure = 5 atm

Solution:

Step 1: Calculate the partial pressure of N2 in air
The mole fraction of N2 in air is given as 0.8. This means that 80% of the air is composed of N2 gas. Since the total mole fraction of all gases in air is 1, the mole fraction of N2 can be converted to its mole fraction in air as follows:

Mole fraction of N2 in air = 0.8
Mole fraction of N2 = 0.8 * 1 = 0.8

Since the mole fraction of N2 in air is 0.8, the partial pressure of N2 in air can be calculated using Dalton's law of partial pressures:

Partial pressure of N2 in air = mole fraction of N2 * total pressure
Partial pressure of N2 in air = 0.8 * 5 atm = 4 atm

Step 2: Calculate the concentration of dissolved N2 in water
Using Henry's Law equation, we can calculate the concentration of dissolved N2 in water:

C = k * P

Substituting the given values:

C = (1.0 x 10^5 atm) * (4 atm) = 4 x 10^5 atm

Step 3: Convert the concentration to moles
The concentration is given in atm, but we need to convert it to moles. To do this, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure in atm
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin

Since the number of moles (n) is the unknown, we can rearrange the equation to solve for n:

n = PV / RT

Given:
P = 4 x 10^5 atm
V = 10 moles (volume of water)
R = 0.0821 L.atm/mol.K
T = 298 K

Substituting the values:

n = (4 x 10^5 atm) * (10 moles) / (0.0821 L.atm/mol.K * 298 K)

Solving this equation will give us the number of moles of N2 dissolved in the water.
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The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 x 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure is. Plz explain how to solve this question. Ans=4×10^-4.plz solve.?
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The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 x 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure is. Plz explain how to solve this question. Ans=4×10^-4.plz solve.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 x 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure is. Plz explain how to solve this question. Ans=4×10^-4.plz solve.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 x 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure is. Plz explain how to solve this question. Ans=4×10^-4.plz solve.?.
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