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The rate of effusion of an unknown gas, X, at 480K is 1.60 times the rate of effusion of sulphur dioxide gas at 300K. The molecular weight of X is ______ (answer should be an integer).
Correct answer is '40'. Can you explain this answer?
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The rate of effusion of an unknown gas, X, at 480K is 1.60 times the r...
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The rate of effusion of an unknown gas, X, at 480K is 1.60 times the r...
Given:
- Rate of effusion of unknown gas X at 480K is 1.60 times the rate of effusion of sulfur dioxide gas at 300K.
To find:
- Molecular weight of gas X.
Explanation:
1. Graham's Law of Effusion:
- Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
2. Mathematical Representation of Graham's Law:
- According to Graham's law, the ratio of the rates of effusion of two gases is given by:
Rate1/Rate2 = √(M2/M1)
Where Rate1 and Rate2 are the rates of effusion of gases 1 and 2, respectively, and M1 and M2 are their respective molar masses.
3. Applying Graham's Law:
- Let's assume the rate of effusion of sulfur dioxide (SO2) gas at 300K is 'R'. Therefore, the rate of effusion of unknown gas X at 480K is 1.60R.
- We can set up the equation:
(1.60R)/R = √(M(SO2)/M(X))
1.60 = √(M(SO2)/M(X))
(1.60)^2 = M(SO2)/M(X)
2.56 = M(SO2)/M(X)
4. Molar Mass of Sulfur Dioxide (SO2):
- The molar mass of sulfur dioxide (SO2) can be calculated as follows:
M(SO2) = (M(S) x 1) + (M(O) x 2)
M(SO2) = 32.07 + (16.00 x 2)
M(SO2) = 32.07 + 32.00
M(SO2) = 64.07 g/mol
5. Solving for M(X):
- Substituting the molar mass of sulfur dioxide (M(SO2) = 64.07 g/mol) and the ratio of effusion rates (2.56) into the equation:
2.56 = 64.07/M(X)
M(X) = 64.07/2.56
M(X) = 25.03 g/mol
Conclusion:
- The molecular weight of gas X is approximately 25.03 g/mol. Since it is required to provide the answer as an integer, the closest whole number is 25. Therefore, the correct answer is 25.
- Rate of effusion of unknown gas X at 480K is 1.60 times the rate of effusion of sulfur dioxide gas at 300K.
To find:
- Molecular weight of gas X.
Explanation:
1. Graham's Law of Effusion:
- Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
2. Mathematical Representation of Graham's Law:
- According to Graham's law, the ratio of the rates of effusion of two gases is given by:
Rate1/Rate2 = √(M2/M1)
Where Rate1 and Rate2 are the rates of effusion of gases 1 and 2, respectively, and M1 and M2 are their respective molar masses.
3. Applying Graham's Law:
- Let's assume the rate of effusion of sulfur dioxide (SO2) gas at 300K is 'R'. Therefore, the rate of effusion of unknown gas X at 480K is 1.60R.
- We can set up the equation:
(1.60R)/R = √(M(SO2)/M(X))
1.60 = √(M(SO2)/M(X))
(1.60)^2 = M(SO2)/M(X)
2.56 = M(SO2)/M(X)
4. Molar Mass of Sulfur Dioxide (SO2):
- The molar mass of sulfur dioxide (SO2) can be calculated as follows:
M(SO2) = (M(S) x 1) + (M(O) x 2)
M(SO2) = 32.07 + (16.00 x 2)
M(SO2) = 32.07 + 32.00
M(SO2) = 64.07 g/mol
5. Solving for M(X):
- Substituting the molar mass of sulfur dioxide (M(SO2) = 64.07 g/mol) and the ratio of effusion rates (2.56) into the equation:
2.56 = 64.07/M(X)
M(X) = 64.07/2.56
M(X) = 25.03 g/mol
Conclusion:
- The molecular weight of gas X is approximately 25.03 g/mol. Since it is required to provide the answer as an integer, the closest whole number is 25. Therefore, the correct answer is 25.
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