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When 2 g of a gas A is introduced into an evacuated flask kept at 25°C, the pressure is found to be one atmosphere. If 3 g of another gas B is then added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal gas behaviour, calculate the ratio of the molecule weights MA : MB.
Correct answer is '1 : 3'. Can you explain this answer?
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°C, the pressure is found to be 0.2 atm. What is the molar mass of gas A?
We can use the ideal gas law to solve for the molar mass of gas A:
PV = nRT
where P is the pressure (in atm), V is the volume (in L), n is the number of moles, R is the gas constant (0.08206 L atm/mol K), and T is the temperature (in K).
First, we need to convert the temperature from Celsius to Kelvin:
T = 25 + 273.15 = 298.15 K
Next, we can solve for the number of moles using the given mass and the molar mass (M) of gas A:
n = m/M
where m is the mass (in g).
We don't know the molar mass yet, so we can rearrange the equation to solve for it:
M = m/n
Substituting the given values, we get:
M = 2 g / (0.2 atm x 0.08206 L atm/mol K x 298.15 K)
M = 45.7 g/mol
Therefore, the molar mass of gas A is approximately 45.7 g/mol.
We can use the ideal gas law to solve for the molar mass of gas A:
PV = nRT
where P is the pressure (in atm), V is the volume (in L), n is the number of moles, R is the gas constant (0.08206 L atm/mol K), and T is the temperature (in K).
First, we need to convert the temperature from Celsius to Kelvin:
T = 25 + 273.15 = 298.15 K
Next, we can solve for the number of moles using the given mass and the molar mass (M) of gas A:
n = m/M
where m is the mass (in g).
We don't know the molar mass yet, so we can rearrange the equation to solve for it:
M = m/n
Substituting the given values, we get:
M = 2 g / (0.2 atm x 0.08206 L atm/mol K x 298.15 K)
M = 45.7 g/mol
Therefore, the molar mass of gas A is approximately 45.7 g/mol.
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When 2 g of a gas A is introduced into an evacuated flask kept at 25°C, the pressure is found to be one atmosphere. If 3 g of another gas B is then added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal gas behaviour, calculate the ratio of the molecule weights MA : MB.Correct answer is '1 : 3'. Can you explain this answer?
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When 2 g of a gas A is introduced into an evacuated flask kept at 25°C, the pressure is found to be one atmosphere. If 3 g of another gas B is then added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal gas behaviour, calculate the ratio of the molecule weights MA : MB.Correct answer is '1 : 3'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about When 2 g of a gas A is introduced into an evacuated flask kept at 25°C, the pressure is found to be one atmosphere. If 3 g of another gas B is then added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal gas behaviour, calculate the ratio of the molecule weights MA : MB.Correct answer is '1 : 3'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When 2 g of a gas A is introduced into an evacuated flask kept at 25°C, the pressure is found to be one atmosphere. If 3 g of another gas B is then added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal gas behaviour, calculate the ratio of the molecule weights MA : MB.Correct answer is '1 : 3'. Can you explain this answer?.
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