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Chapter 10: Problem 92

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 105 s for \(1.0 \mathrm{~L}\) of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of \(1.0 \mathrm{~L} ;\) that is, rate and time are inversely proportional.)

Short Answer

Expert verified
The molar mass of the unknown gas is approximately 9.96 g/mol.

Step by step solution

01

Write the formula for Graham's Law of Effusion in terms of time and molar mass

Given that rate and time are inversely proportional, we can rewrite Graham's Law of Effusion as the ratio of the square root of the molar masses instead of the effusion rates. \[ \frac{\sqrt{MM_{1}}}{\sqrt{MM_{2}}} = \frac{t_{2}}{t_{1}} \]Where: - \(MM_{1}\) is the molar mass of the unknown gas - \(MM_{2}\) is the molar mass of O₂ gas - \(t_{1}\) is the time for the unknown gas to effuse (105 s) - \(t_{2}\) is the time for the O₂ gas to effuse (31 s)
02

Insert the given values and rearrange the formula to solve for \(MM_{1}\)

We know the molar mass of O₂ gas is 32 g/mol. Plugging in the given values into the formula, we get:\[ \frac{\sqrt{MM_{1}}}{\sqrt{32}} = \frac{31}{105} \]Now, we will rearrange the formula to solve for the molar mass of the unknown gas, \(MM_{1}\):\[ \sqrt{MM_{1}} = \frac{31}{105} \times \sqrt{32} \]Square both sides of the equation to get rid of the square root:\[ MM_{1} = \left(\frac{31}{105} \times \sqrt{32}\right)^2 \]
03

Calculate the molar mass of the unknown gas

Now, we can calculate the value of \(MM_{1}\) using the rearranged formula:\[ MM_{1} = \left(\frac{31}{105} \times \sqrt{32}\right)^2 \Rightarrow MM_{1} \approx 9.96 \] So, the molar mass of the unknown gas is approximately 9.96 g/mol.

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