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A vessel contains a mixture of H2 and N2 gas. The density of this gas mixture is 0.2 g L–1 at 300 K and 1 atm. Assuming that both the gases behave ideally, the mole fraction of N2(g) in the vessel is __________.
(final answer should be rounded off to two decimal places)
[Given: R = 0.082 L atm mol–1 K–1, atomic wt. of hydrogen = 1.0 and atomic wt. of nitrogen = 14.0]
(final answer should be rounded off to two decimal places)
[Given: R = 0.082 L atm mol–1 K–1, atomic wt. of hydrogen = 1.0 and atomic wt. of nitrogen = 14.0]
Correct answer is '0.11'. Can you explain this answer?
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A vessel contains a mixture of H2 and N2 gas. The density of this gas ...
Solution:
Given data:
Density of gas mixture = 0.2 g L^-1
Temperature (T) = 300 K
Pressure (P) = 1 atm
R = 0.082 L atm mol^-1 K^-1
Atomic weight of hydrogen (H) = 1.0
Atomic weight of nitrogen (N) = 14.0
We need to find the mole fraction of N2 gas in the vessel.
Step 1: Calculate the total number of moles of gases in the vessel
We know that the density (d) of a gas is related to its molar mass (M), pressure (P) and temperature (T) by the following equation:
d = PM/RT
Rearranging the equation, we get:
M = dRT/P
Substituting the given values, we get:
M = 0.2 g L^-1 x 0.082 L atm mol^-1 K^-1 x 300 K / 1 atm
M = 4.92 g mol^-1
Let the number of moles of H2 gas in the vessel be n1 and that of N2 gas be n2. Then:
n1 + n2 = total number of moles of gases in the vessel = M/atomic weight of the mixture
The atomic weight of the mixture is given by:
atomic weight of the mixture = (1 x atomic weight of H2 + 1 x atomic weight of N2)/2
= (1 x 1.0 + 1 x 14.0)/2
= 7.5
Substituting the values, we get:
n1 + n2 = 4.92 g mol^-1 / 7.5 g mol^-1
n1 + n2 = 0.656 mol
Step 2: Calculate the partial pressures of H2 and N2 gases in the vessel
According to Dalton's law of partial pressures, the total pressure of a gas mixture is equal to the sum of the partial pressures of the component gases. Therefore:
P(total) = P(H2) + P(N2)
The partial pressure of a gas is related to its mole fraction (x) and the total pressure (P) by the following equation:
P(gas) = x(gas) x P(total)
Let x(N2) be the mole fraction of N2 gas in the vessel. Then:
x(H2) = 1 - x(N2)
Substituting the values, we get:
P(H2) = (1 - x(N2)) x 1 atm = (1 - x(N2)) atm
P(N2) = x(N2) x 1 atm = x(N2) atm
Step 3: Calculate the mole fraction of N2 gas in the vessel
The total pressure of the gas mixture is 1 atm. Therefore:
P(total) = P(H2) + P(N2) = (1 - x(N2)) atm + x(N2) atm = 1 atm
Substituting the values, we get:
(1 - x(N2)) + x(N2) = 1
1 - x(N2) + x(N2) = 1
x(N2) = 0.11 (rounded off to two decimal places)
Therefore, the mole fraction of N2 gas in
Given data:
Density of gas mixture = 0.2 g L^-1
Temperature (T) = 300 K
Pressure (P) = 1 atm
R = 0.082 L atm mol^-1 K^-1
Atomic weight of hydrogen (H) = 1.0
Atomic weight of nitrogen (N) = 14.0
We need to find the mole fraction of N2 gas in the vessel.
Step 1: Calculate the total number of moles of gases in the vessel
We know that the density (d) of a gas is related to its molar mass (M), pressure (P) and temperature (T) by the following equation:
d = PM/RT
Rearranging the equation, we get:
M = dRT/P
Substituting the given values, we get:
M = 0.2 g L^-1 x 0.082 L atm mol^-1 K^-1 x 300 K / 1 atm
M = 4.92 g mol^-1
Let the number of moles of H2 gas in the vessel be n1 and that of N2 gas be n2. Then:
n1 + n2 = total number of moles of gases in the vessel = M/atomic weight of the mixture
The atomic weight of the mixture is given by:
atomic weight of the mixture = (1 x atomic weight of H2 + 1 x atomic weight of N2)/2
= (1 x 1.0 + 1 x 14.0)/2
= 7.5
Substituting the values, we get:
n1 + n2 = 4.92 g mol^-1 / 7.5 g mol^-1
n1 + n2 = 0.656 mol
Step 2: Calculate the partial pressures of H2 and N2 gases in the vessel
According to Dalton's law of partial pressures, the total pressure of a gas mixture is equal to the sum of the partial pressures of the component gases. Therefore:
P(total) = P(H2) + P(N2)
The partial pressure of a gas is related to its mole fraction (x) and the total pressure (P) by the following equation:
P(gas) = x(gas) x P(total)
Let x(N2) be the mole fraction of N2 gas in the vessel. Then:
x(H2) = 1 - x(N2)
Substituting the values, we get:
P(H2) = (1 - x(N2)) x 1 atm = (1 - x(N2)) atm
P(N2) = x(N2) x 1 atm = x(N2) atm
Step 3: Calculate the mole fraction of N2 gas in the vessel
The total pressure of the gas mixture is 1 atm. Therefore:
P(total) = P(H2) + P(N2) = (1 - x(N2)) atm + x(N2) atm = 1 atm
Substituting the values, we get:
(1 - x(N2)) + x(N2) = 1
1 - x(N2) + x(N2) = 1
x(N2) = 0.11 (rounded off to two decimal places)
Therefore, the mole fraction of N2 gas in
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A vessel contains a mixture of H2 and N2 gas. The density of this gas mixture is 0.2 g L1 at 300 K and 1 atm. Assuming that both the gases behave ideally, the mole fraction of N2(g) in the vessel is __________.(final answer should be rounded off to two decimal places)[Given: R = 0.082 L atm mol1 K1, atomic wt. of hydrogen = 1.0 and atomic wt. of nitrogen = 14.0]Correct answer is '0.11'. Can you explain this answer?
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A vessel contains a mixture of H2 and N2 gas. The density of this gas mixture is 0.2 g L1 at 300 K and 1 atm. Assuming that both the gases behave ideally, the mole fraction of N2(g) in the vessel is __________.(final answer should be rounded off to two decimal places)[Given: R = 0.082 L atm mol1 K1, atomic wt. of hydrogen = 1.0 and atomic wt. of nitrogen = 14.0]Correct answer is '0.11'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A vessel contains a mixture of H2 and N2 gas. The density of this gas mixture is 0.2 g L1 at 300 K and 1 atm. Assuming that both the gases behave ideally, the mole fraction of N2(g) in the vessel is __________.(final answer should be rounded off to two decimal places)[Given: R = 0.082 L atm mol1 K1, atomic wt. of hydrogen = 1.0 and atomic wt. of nitrogen = 14.0]Correct answer is '0.11'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A vessel contains a mixture of H2 and N2 gas. The density of this gas mixture is 0.2 g L1 at 300 K and 1 atm. Assuming that both the gases behave ideally, the mole fraction of N2(g) in the vessel is __________.(final answer should be rounded off to two decimal places)[Given: R = 0.082 L atm mol1 K1, atomic wt. of hydrogen = 1.0 and atomic wt. of nitrogen = 14.0]Correct answer is '0.11'. Can you explain this answer?.
A vessel contains a mixture of H2 and N2 gas. The density of this gas mixture is 0.2 g L1 at 300 K and 1 atm. Assuming that both the gases behave ideally, the mole fraction of N2(g) in the vessel is __________.(final answer should be rounded off to two decimal places)[Given: R = 0.082 L atm mol1 K1, atomic wt. of hydrogen = 1.0 and atomic wt. of nitrogen = 14.0]Correct answer is '0.11'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A vessel contains a mixture of H2 and N2 gas. The density of this gas mixture is 0.2 g L1 at 300 K and 1 atm. Assuming that both the gases behave ideally, the mole fraction of N2(g) in the vessel is __________.(final answer should be rounded off to two decimal places)[Given: R = 0.082 L atm mol1 K1, atomic wt. of hydrogen = 1.0 and atomic wt. of nitrogen = 14.0]Correct answer is '0.11'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A vessel contains a mixture of H2 and N2 gas. The density of this gas mixture is 0.2 g L1 at 300 K and 1 atm. Assuming that both the gases behave ideally, the mole fraction of N2(g) in the vessel is __________.(final answer should be rounded off to two decimal places)[Given: R = 0.082 L atm mol1 K1, atomic wt. of hydrogen = 1.0 and atomic wt. of nitrogen = 14.0]Correct answer is '0.11'. Can you explain this answer?.
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A vessel contains a mixture of H2 and N2 gas. The density of this gas mixture is 0.2 g L1 at 300 K and 1 atm. Assuming that both the gases behave ideally, the mole fraction of N2(g) in the vessel is __________.(final answer should be rounded off to two decimal places)[Given: R = 0.082 L atm mol1 K1, atomic wt. of hydrogen = 1.0 and atomic wt. of nitrogen = 14.0]Correct answer is '0.11'. Can you explain this answer?, a detailed solution for A vessel contains a mixture of H2 and N2 gas. The density of this gas mixture is 0.2 g L1 at 300 K and 1 atm. Assuming that both the gases behave ideally, the mole fraction of N2(g) in the vessel is __________.(final answer should be rounded off to two decimal places)[Given: R = 0.082 L atm mol1 K1, atomic wt. of hydrogen = 1.0 and atomic wt. of nitrogen = 14.0]Correct answer is '0.11'. Can you explain this answer? has been provided alongside types of A vessel contains a mixture of H2 and N2 gas. The density of this gas mixture is 0.2 g L1 at 300 K and 1 atm. Assuming that both the gases behave ideally, the mole fraction of N2(g) in the vessel is __________.(final answer should be rounded off to two decimal places)[Given: R = 0.082 L atm mol1 K1, atomic wt. of hydrogen = 1.0 and atomic wt. of nitrogen = 14.0]Correct answer is '0.11'. Can you explain this answer? theory, EduRev gives you an
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