## Related questions with answers

Question

What is the rate of effusion for a gas that has a molar mass twice that of a gas that effuses at a rate of 3.6 mol/min?

Solutions

VerifiedSolution A

Solution B

Answered 11 months ago

Step 1

1 of 2Answered 2 years ago

Step 1

1 of 2Rate formula:

$\dfrac{Rate_{1}}{Rate_{2}}=\sqrt{\dfrac{M_{2}}{M_{1}}}$

Known data:

$Rate_{1}= 3.6 \text{mol/min}$

$M_{2}=2\times M_{1}$

Therefore:

$\dfrac{3.6\text{mol/min}}{Rate_{2}}=\sqrt{\dfrac{2\times M_{1}}{M_{1}}}$

$3.6\text{mol/min}=Rate_{2}\times\sqrt{2}$

$Rate_{2}=\dfrac{3.6\text{mol/min}}{\sqrt{2}}=2.55\text{mol/min}$

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