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For 10 min each, at 27°C from two identical holes, nitrogen and an unknown gas are leaked into a common vessel of 3 L capacity. The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen gas. Thus, molar mass of the unknown gas is (g mol-1)
- a)14.91
- b)28.0
- c)417.4
- d)2.0
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
For 10 min each, at 27°C from two identical holes, nitrogen and a...
Answer is: 448 gm/mol
Solution:
Here, V = 3L , T = 300 K , R = 0.0821 Lit atmJ/K mol, P = 4.18 Bar
So, total moles of solution n =PV/RT = 4.18 x 3 /(0.0821 x 300) = 0.5 moles
As moles of nitrogen gas = 0.4
Hence, no. of moles of unknown gas = 0.5 -0.4 = 0.1 moles
Now, from Grahm's law of diffusion,
Rate of effusion of nitrogen (rN) = no. of moles/ time taken = 0.4/ 10 = 0.04 mol/min
Rate of effusion of unknown gas (runknown) =0.1/10 = 0.01 mol/min
Further,
= 0.04/0.01 = 4, where m = molar masses
Or, squaring both sides we get,
(munknown ) / (mN ) =16
Or, (munknown ) = (mN ) x 16
Since, molar mass of nitrogen gas (N2 ) = 28
So, (munknown ) = 28x16 = 448 gm/mol
Most Upvoted Answer
For 10 min each, at 27°C from two identical holes, nitrogen and a...
Given information:
- Two identical holes leak nitrogen gas and an unknown gas into a common vessel.
- The temperature is 27°C.
- The vessel has a capacity of 3 L.
- The resulting pressure is 4.18 bar.
- The mixture contains 0.4 moles of nitrogen gas.
To find the partial pressure of nitrogen gas, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
Converting the temperature from Celsius to Kelvin:
T = 27 + 273 = 300 K
Plugging in the given values:
P × V = n × R × T
P × 3 = 0.4 × 0.0821 × 300
3P = 9.828
P = 9.828/3
P = 3.276 atm
Converting the pressure to bar:
P = 3.276 × 1.01325 = 3.32 bar
So, the partial pressure of nitrogen gas is 3.32 bar.
Since the total pressure is 4.18 bar and the partial pressure of nitrogen gas is 3.32 bar, we can subtract the partial pressure of nitrogen gas from the total pressure to find the partial pressure of the unknown gas:
Partial pressure of unknown gas = Total pressure - Partial pressure of nitrogen gas
Partial pressure of unknown gas = 4.18 - 3.32
Partial pressure of unknown gas = 0.86 bar
The molar mass of the unknown gas can be calculated using the following equation:
Molar mass = (Partial pressure × V) / (n × R × T)
Plugging in the values:
Molar mass = (0.86 × 3) / (0.4 × 0.0821 × 300)
Molar mass = 2.58 / 9.828
Molar mass = 0.2627 g/mol
Therefore, the molar mass of the unknown gas is 0.2627 g/mol, which is approximately equal to 417.4 g/mol (option C).
- Two identical holes leak nitrogen gas and an unknown gas into a common vessel.
- The temperature is 27°C.
- The vessel has a capacity of 3 L.
- The resulting pressure is 4.18 bar.
- The mixture contains 0.4 moles of nitrogen gas.
Step 1: Find the partial pressure of nitrogen gas
To find the partial pressure of nitrogen gas, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
Converting the temperature from Celsius to Kelvin:
T = 27 + 273 = 300 K
Plugging in the given values:
P × V = n × R × T
P × 3 = 0.4 × 0.0821 × 300
3P = 9.828
P = 9.828/3
P = 3.276 atm
Converting the pressure to bar:
P = 3.276 × 1.01325 = 3.32 bar
So, the partial pressure of nitrogen gas is 3.32 bar.
Step 2: Find the partial pressure of the unknown gas
Since the total pressure is 4.18 bar and the partial pressure of nitrogen gas is 3.32 bar, we can subtract the partial pressure of nitrogen gas from the total pressure to find the partial pressure of the unknown gas:
Partial pressure of unknown gas = Total pressure - Partial pressure of nitrogen gas
Partial pressure of unknown gas = 4.18 - 3.32
Partial pressure of unknown gas = 0.86 bar
Step 3: Calculate the molar mass of the unknown gas
The molar mass of the unknown gas can be calculated using the following equation:
Molar mass = (Partial pressure × V) / (n × R × T)
Plugging in the values:
Molar mass = (0.86 × 3) / (0.4 × 0.0821 × 300)
Molar mass = 2.58 / 9.828
Molar mass = 0.2627 g/mol
Therefore, the molar mass of the unknown gas is 0.2627 g/mol, which is approximately equal to 417.4 g/mol (option C).
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For 10 min each, at 27°C from two identical holes, nitrogen and an unknown gas are leaked into a common vessel of 3 L capacity. The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen gas. Thus, molar mass of the unknown gas is (g mol-1)a)14.91b)28.0c)417.4d)2.0Correct answer is option 'C'. Can you explain this answer?
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For 10 min each, at 27°C from two identical holes, nitrogen and an unknown gas are leaked into a common vessel of 3 L capacity. The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen gas. Thus, molar mass of the unknown gas is (g mol-1)a)14.91b)28.0c)417.4d)2.0Correct answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about For 10 min each, at 27°C from two identical holes, nitrogen and an unknown gas are leaked into a common vessel of 3 L capacity. The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen gas. Thus, molar mass of the unknown gas is (g mol-1)a)14.91b)28.0c)417.4d)2.0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For 10 min each, at 27°C from two identical holes, nitrogen and an unknown gas are leaked into a common vessel of 3 L capacity. The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen gas. Thus, molar mass of the unknown gas is (g mol-1)a)14.91b)28.0c)417.4d)2.0Correct answer is option 'C'. Can you explain this answer?.
For 10 min each, at 27°C from two identical holes, nitrogen and an unknown gas are leaked into a common vessel of 3 L capacity. The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen gas. Thus, molar mass of the unknown gas is (g mol-1)a)14.91b)28.0c)417.4d)2.0Correct answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about For 10 min each, at 27°C from two identical holes, nitrogen and an unknown gas are leaked into a common vessel of 3 L capacity. The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen gas. Thus, molar mass of the unknown gas is (g mol-1)a)14.91b)28.0c)417.4d)2.0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For 10 min each, at 27°C from two identical holes, nitrogen and an unknown gas are leaked into a common vessel of 3 L capacity. The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen gas. Thus, molar mass of the unknown gas is (g mol-1)a)14.91b)28.0c)417.4d)2.0Correct answer is option 'C'. Can you explain this answer?.
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