2. An unknown gas diffuses 1.5 times slower than nitrogen, N2. If the molar mass of nitrogen is \( 28 \mathrm{~g} / \mathrm{mol} \), find the molar mass of the unknown gas.
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
Explain 
Simplify this solution Beyond the Answer
To solve for the molar mass of the unknown gas, we can use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. If the unknown gas diffuses 1.5 times slower than nitrogen (N₂), we can set up the equation: \[ \frac{\text{Rate of N}_2}{\text{Rate of unknown gas}} = \sqrt{\frac{M_{\text{unknown}}}{M_{N2}}} \] Given the rate ratio of 1:1.5, we rewrite it as: \[ \frac{1}{1.5} = \sqrt{\frac{M_{\text{unknown}}}{28}} \] Squaring both sides gives us: \[ \left(\frac{1}{1.5}\right)^2 = \frac{M_{\text{unknown}}}{28} \] Calculating \(\left(\frac{1}{1.5}\right)^2\) results in \(\frac{1}{2.25} = \frac{28}{M_{\text{unknown}}}\). Now, cross-multiply and solve for \(M_{\text{unknown}}\): \[ M_{\text{unknown}} = 28 \times 2.25 = 63 \mathrm{~g/mol} \] So, the molar mass of the unknown gas is \(63 \mathrm{~g/mol}\). --- During your next trip on an aircraft, you can marvel at how the principles of gas diffusion impact cabin pressure and oxygen levels! Understanding diffusion rates is critical for aerospace engineers to design systems that ensure optimal performance and safety during high-altitude flights. If you're analyzing the properties of various gases, remember to consider temperature and pressure conditions! Common mistakes include overlooking the effect of these variables, which can lead to inaccurate molecular weight calculations. Always ensure that you’re working under ideal gas conditions when making such assessments!
