Interpretation:
The rate of effusion for an unknown gas, having a molar mass twice the gas that effuses at a rate of 3.6 mol/min needs to be determined.
Concept introduction:
According to the Graham’s law of effusion, rate of effusion of any gas is always inversely proportional to the square root of the mass of gaseous particles. The mathematical expression for this law is:
In terms of time for gases; this can be written as:
Answer to Problem 3PP
Explanation of Solution
Given information :
Rate of effusion for gas-2 = 3.6 mol/min
Molar mass of gas-1 = 2 x molar mass of gas-1
Putting the values in the equation:
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Chemistry: Matter and Change
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