Question

a sample of n2 gas (2.0 mmol) effused through a pinhole in 5.5 s. how long will it take for the same amount of ch4 to effuse under the same conditions?

Answer 1

By using Graham's law we can calculate that it will take the methane (CH₄) sample 4.2 s to effuse under the same conditions.

Graham's law connects the rates of effusion (RoE) of two gases and their molar masses (M):

$\frac{RoE(A)}{RoE(B)} = \sqrt{\frac{M(B)}{M(A)} }$

We can calculate the RoE for N₂ by using the given number of moles (n = 2.0 mol) and time (t = 5.5 s) needed for it to effuse:

RoE(N₂) = n/t

RoE(N₂) = 2.0 mmol / 5.5 s

RoE(N₂) = 0.36 mmol/s

Now, we can use the molar masses of nitrogen (M = 28 g/mol) and methane (M = 16 g/mol) to calculate the RoE(CH₄):

$\frac{RoE(N_{2} )}{RoE(CH_{4} )} = \sqrt{\frac{M(CH_{4})}{M(N_{2} )} }$

$RoE(CH_{4} ) = \frac{RoE(N_{2} )}{\sqrt{\frac{M(CH_{4} )}{M(N_{2} )} } }$

$RoE(CH_{4} ) = \frac{0.36 mmol/s}{\sqrt{\frac{16 g/mol}{28 g/mol} } }$

RoE(CH₄) = 0.48 mmol/s

Now we can use this to calculate the time 2.0 mmol of methane will require:

t = n(CH₄) / RoE(CH₄)

t = 2.0 mmol / 0.48 mmol/s

t = 4.2 s

You can learn more about Graham's law here:

brainly.com/question/12415336

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