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Chapter 5: Problem 104

The rate of effusion of a particular gas was measured and found to be \(24.0 \mathrm{~mL} / \mathrm{min}\). Under the same conditions, the rate of effusion of pure methane \(\left(\mathrm{CH}_{4}\right)\) gas is \(47.8 \mathrm{~mL} / \mathrm{min}\). What is the molar mass of the unknown gas?

Short Answer

Expert verified
The molar mass of the unknown gas is approximately \(44.493 \frac{g}{mol}\).

Step by step solution

01

Write the formula for Graham's Law of Effusion

Graham's Law of Effusion is given by the formula: \[ \frac{Rate_1}{Rate_2} = \sqrt{\frac{M_2}{M_1}} \] Where \(Rate_1\) and \(Rate_2\) are the rates of effusion of two gases, and \(M_1\) and \(M_2\) are their respective molar masses.
02

Input the given values into the formula

In this problem, we're given the rate of effusion of the unknown gas (\(Rate_1\)), the rate of effusion of methane (\(Rate_2\)), and the molar mass of methane (\(M_2\)). The molar mass of methane is \(16.04 \frac{g}{mol}\). Let's plug these values into the formula: \[ \frac{24.0 mL/min}{47.8 mL/min} = \sqrt{\frac{16.04 \frac{g}{mol}}{M_1}} \]
03

Solve for M1, the molar mass of the unknown gas

In order to solve for \(M_1\), we have to simplify and rearrange the equation as follows: \[ \frac{24.0}{47.8} = \sqrt{\frac{16.04}{M_1}} \] Now, square both sides of the equation: \[ \left(\frac{24.0}{47.8}\right)^2 = \frac{16.04}{M_1} \] Next, solve for \(M_1\) by multiplying both sides of the equation by \(M_1\) and then dividing both sides by \(\left(\frac{24.0}{47.8}\right)^2\): \[ M_1 = \frac{16.04}{\left(\frac{24.0}{47.8}\right)^2} \] Finally, calculate \(M_1\): \[ M_1 \approx 44.493 \frac{g}{mol}\]
04

State the answer

The molar mass of the unknown gas is approximately \(44.493 \frac{g}{mol}\).

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