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A vessel contains a mixture of H2 and N2 gas. The density of the gas mixture is 0.2g/L at 300K and 1atm. 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=0.082L.atm.mol^-1.K^-1, atomic weight of hydrogen=1 and atomic weight of nitrogen=14]?
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A vessel contains a mixture of H2 and N2 gas. The density of the gas m...
Ans.
Given: density of mixture=0.2g/L ,T=300K,P=1atm
Using formula d=PM/RT
Hence, M= (dxRxT)/P
M=(0.2x0.0821x300)/1
M=4.926g
Now, let mole fraction of N2 (X1) be 'n' and mole fraction of H2(X2) be '(1-n)'
M1=( 2x14=28)=mol.mass of N2
M2=(2x1=2)=mol.mass of H2
Applying formula of molar mass of mixture—
M= M1 X1 +M2 X2
4.926={ 28xn + 2x(1-n) }
solving for ‘n’ we find n=0.11
Hence the mole fraction of N2 (g)in the vessel is 0.11
Using formula d=PM/RT
Hence, M= (dxRxT)/P
M=(0.2x0.0821x300)/1
M=4.926g
Now, let mole fraction of N2 (X1) be 'n' and mole fraction of H2(X2) be '(1-n)'
M1=( 2x14=28)=mol.mass of N2
M2=(2x1=2)=mol.mass of H2
Applying formula of molar mass of mixture—
M= M1 X1 +M2 X2
4.926={ 28xn + 2x(1-n) }
solving for ‘n’ we find n=0.11
Hence the mole fraction of N2 (g)in the vessel is 0.11
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A vessel contains a mixture of H2 and N2 gas. The density of the gas m...
Given:
- Density of the gas mixture = 0.2 g/L
- Temperature (T) = 300 K
- Pressure (P) = 1 atm
- R = 0.082 L.atm.mol^-1.K^-1 (Ideal Gas Constant)
- Atomic weight of hydrogen (H2) = 1
- Atomic weight of nitrogen (N2) = 14
Calculating the Molar Mass of the Gas Mixture:
The molar mass of a gas mixture can be calculated using the ideal gas equation:
PV = nRT
Where:
- P is the pressure of the gas mixture
- V is the volume of the gas mixture
- n is the number of moles of the gas mixture
- R is the ideal gas constant
- T is the temperature of the gas mixture
Rearranging the equation to solve for n:
n = PV / RT
The molar mass (M) of the gas mixture can be calculated using the formula:
M = mass / n
Given that the density (d) of the gas mixture is given in grams per liter, we can convert it to grams per mole by multiplying by the molar volume (Vm):
d = mass / Vm
Rearranging the equation to solve for mass:
mass = d * Vm
Substituting the value of mass into the molar mass formula:
M = (d * Vm) / n
Calculating the Mole Fraction of N2:
The mole fraction (X) of a gas component in a mixture is the ratio of the number of moles of that component to the total number of moles in the mixture.
For a binary mixture of H2 and N2, the mole fraction of N2 (X_N2) can be calculated using the formula:
X_N2 = n_N2 / (n_H2 + n_N2)
Since we know the molar mass of the gas mixture, we can calculate the number of moles of N2 (n_N2) using the formula:
n_N2 = mass_N2 / M_N2
Similarly, we can calculate the number of moles of H2 (n_H2) using the formula:
n_H2 = mass_H2 / M_H2
Substituting the values into the mole fraction formula:
X_N2 = (mass_N2 / M_N2) / ((mass_H2 / M_H2) + (mass_N2 / M_N2))
Solving the Problem:
1. Calculate the molar mass of the gas mixture using the given density:
- mass = d * Vm
- M = (d * Vm) / n
2. Calculate the number of moles of N2 and H2 using their molar masses:
- n_N2 = mass_N2 / M_N2
- n_H2 = mass_H2 / M_H2
3. Calculate the mole fraction of N2 in the gas mixture:
- X_N2 = (n_N2) / (n_H2 + n_N2)
4. Round the final answer to two decimal places.
By following the above steps and substituting the given values, you can calculate the mole fraction of N2 in the gas mixture.
- Density of the gas mixture = 0.2 g/L
- Temperature (T) = 300 K
- Pressure (P) = 1 atm
- R = 0.082 L.atm.mol^-1.K^-1 (Ideal Gas Constant)
- Atomic weight of hydrogen (H2) = 1
- Atomic weight of nitrogen (N2) = 14
Calculating the Molar Mass of the Gas Mixture:
The molar mass of a gas mixture can be calculated using the ideal gas equation:
PV = nRT
Where:
- P is the pressure of the gas mixture
- V is the volume of the gas mixture
- n is the number of moles of the gas mixture
- R is the ideal gas constant
- T is the temperature of the gas mixture
Rearranging the equation to solve for n:
n = PV / RT
The molar mass (M) of the gas mixture can be calculated using the formula:
M = mass / n
Given that the density (d) of the gas mixture is given in grams per liter, we can convert it to grams per mole by multiplying by the molar volume (Vm):
d = mass / Vm
Rearranging the equation to solve for mass:
mass = d * Vm
Substituting the value of mass into the molar mass formula:
M = (d * Vm) / n
Calculating the Mole Fraction of N2:
The mole fraction (X) of a gas component in a mixture is the ratio of the number of moles of that component to the total number of moles in the mixture.
For a binary mixture of H2 and N2, the mole fraction of N2 (X_N2) can be calculated using the formula:
X_N2 = n_N2 / (n_H2 + n_N2)
Since we know the molar mass of the gas mixture, we can calculate the number of moles of N2 (n_N2) using the formula:
n_N2 = mass_N2 / M_N2
Similarly, we can calculate the number of moles of H2 (n_H2) using the formula:
n_H2 = mass_H2 / M_H2
Substituting the values into the mole fraction formula:
X_N2 = (mass_N2 / M_N2) / ((mass_H2 / M_H2) + (mass_N2 / M_N2))
Solving the Problem:
1. Calculate the molar mass of the gas mixture using the given density:
- mass = d * Vm
- M = (d * Vm) / n
2. Calculate the number of moles of N2 and H2 using their molar masses:
- n_N2 = mass_N2 / M_N2
- n_H2 = mass_H2 / M_H2
3. Calculate the mole fraction of N2 in the gas mixture:
- X_N2 = (n_N2) / (n_H2 + n_N2)
4. Round the final answer to two decimal places.
By following the above steps and substituting the given values, you can calculate the mole fraction of N2 in the gas mixture.
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A vessel contains a mixture of H2 and N2 gas. The density of the gas mixture is 0.2g/L at 300K and 1atm. 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=0.082L.atm.mol^-1.K^-1, atomic weight of hydrogen=1 and atomic weight of nitrogen=14]?
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A vessel contains a mixture of H2 and N2 gas. The density of the gas mixture is 0.2g/L at 300K and 1atm. 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=0.082L.atm.mol^-1.K^-1, atomic weight of hydrogen=1 and atomic weight of nitrogen=14]? 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 the gas mixture is 0.2g/L at 300K and 1atm. 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=0.082L.atm.mol^-1.K^-1, atomic weight of hydrogen=1 and atomic weight of nitrogen=14]? 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 the gas mixture is 0.2g/L at 300K and 1atm. 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=0.082L.atm.mol^-1.K^-1, atomic weight of hydrogen=1 and atomic weight of nitrogen=14]?.
A vessel contains a mixture of H2 and N2 gas. The density of the gas mixture is 0.2g/L at 300K and 1atm. 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=0.082L.atm.mol^-1.K^-1, atomic weight of hydrogen=1 and atomic weight of nitrogen=14]? 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 the gas mixture is 0.2g/L at 300K and 1atm. 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=0.082L.atm.mol^-1.K^-1, atomic weight of hydrogen=1 and atomic weight of nitrogen=14]? 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 the gas mixture is 0.2g/L at 300K and 1atm. 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=0.082L.atm.mol^-1.K^-1, atomic weight of hydrogen=1 and atomic weight of nitrogen=14]?.
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A vessel contains a mixture of H2 and N2 gas. The density of the gas mixture is 0.2g/L at 300K and 1atm. 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=0.082L.atm.mol^-1.K^-1, atomic weight of hydrogen=1 and atomic weight of nitrogen=14]?, a detailed solution for A vessel contains a mixture of H2 and N2 gas. The density of the gas mixture is 0.2g/L at 300K and 1atm. 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=0.082L.atm.mol^-1.K^-1, atomic weight of hydrogen=1 and atomic weight of nitrogen=14]? has been provided alongside types of A vessel contains a mixture of H2 and N2 gas. The density of the gas mixture is 0.2g/L at 300K and 1atm. 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=0.082L.atm.mol^-1.K^-1, atomic weight of hydrogen=1 and atomic weight of nitrogen=14]? theory, EduRev gives you an
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