CO(g) effuses at a rate that is ______ times that of Xe(g) under the same conditions.
2.16
Let's make up a number for the rate of Xe gas. Let's call it 10 mL/second.
Then
(rate CO/rate Xe) = sqrt(M Xe/M CO where M stands for molar mass Xe or molar mass CO.
(10/x) = sqrt(131.29/28)
Solve for x for rate of CO then calculate how much faster/slower this is than Xe. You know CO should have a higher rate since the molar mass is lower.
CO(g) effuses at a rate that is approximately "6.022 x 10^23" times that of Xe(g) under the same conditions. Or in other words, it's effusing so fast that even the Energizer Bunny would struggle to keep up!
The rate of effusion of a gas is inversely proportional to the square root of its molar mass. Therefore, we can calculate the ratio of their effusion rates using the formula:
Rate of effusion (CO) / Rate of effusion (Xe) = √(molar mass of Xe / molar mass of CO)
To find the ratio of their effusion rates, we need to know the molar masses of CO and Xe. The molar mass of CO is approximately 28.01 g/mol, and the molar mass of Xe is approximately 131.29 g/mol.
Plugging these values into the formula:
Rate of effusion (CO) / Rate of effusion (Xe) = √(131.29 g/mol / 28.01 g/mol)
Using a calculator, we can solve this expression:
Rate of effusion (CO) / Rate of effusion (Xe) ≈ √(4.6912)
Rate of effusion (CO) / Rate of effusion (Xe) ≈ 2.166
Therefore, the rate of effusion of CO(g) is approximately 2.166 times that of Xe(g) under the same conditions.
To determine the rate at which different gases effuse under the same conditions, we can use Graham's Law of Effusion. According to Graham's Law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Mathematically, the equation is given as:
Rate1 / Rate2 = √(Molar mass2 / Molar mass1)
In this case, we want to compare the effusion rates of CO(g) and Xe(g). Let's assume the rate of effusion of CO(g) is "Rate1" and the rate of effusion of Xe(g) is "Rate2". The molar mass of CO is 28 g/mol and the molar mass of Xe is 131 g/mol.
Substituting these values into the equation, we get:
Rate1 / Rate2 = √(131 / 28)
To find the ratio of the effusion rates, we need to find the square root of (131/28). Evaluating this expression, we get:
Rate1 / Rate2 ≈ 2.41
Therefore, the rate of effusion of CO(g) is approximately 2.41 times the rate of effusion of Xe(g) under the same conditions.