Use Graham's law of effusion to calculate molar mass. A sample of krypton, Kr, effuses through a small hole at a rate of 9.00×10-6 mol/h. An unknown gas, under the same conditions, effuses at a rate of 1.93×105 mol/h. Calculate the molar mass of the unknown gas. g/mol
Use Graham's law of effusion to calculate molar mass. A sample of krypton, Kr, effuses through a small hole at a rate of 9.00×10-6 mol/h. An unknown gas, under the same conditions, effuses at a rate of 1.93×105 mol/h. Calculate the molar mass of the unknown gas. g/mol
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Chapter1: Chemical Foundations
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![**Using Graham's Law of Effusion to Calculate Molar Mass**
A sample of **krypton**, Kr, effuses through a small hole at a rate of \(9.00 \times 10^{-6}\) mol/h. An unknown gas, under the same conditions, effuses at a rate of \(1.93 \times 10^{-5}\) mol/h. Calculate the molar mass of the unknown gas.
\[ \_\_\_\_\_\_ \text{ g/mol} \]
(Note: In order to find the molar mass of the unknown gas, you can apply 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. The formula is \(\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}}\), where \(\text{Rate}_1\) and \(\text{Rate}_2\) are the effusion rates of the gases, and \(M_1\) and \(M_2\) are their molar masses.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb8731af5-23ed-4fe5-8b12-d0cb96af1906%2Feb786cb1-dcae-46db-bb3e-b689524a0885%2Frwge1ho_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Using Graham's Law of Effusion to Calculate Molar Mass**
A sample of **krypton**, Kr, effuses through a small hole at a rate of \(9.00 \times 10^{-6}\) mol/h. An unknown gas, under the same conditions, effuses at a rate of \(1.93 \times 10^{-5}\) mol/h. Calculate the molar mass of the unknown gas.
\[ \_\_\_\_\_\_ \text{ g/mol} \]
(Note: In order to find the molar mass of the unknown gas, you can apply 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. The formula is \(\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}}\), where \(\text{Rate}_1\) and \(\text{Rate}_2\) are the effusion rates of the gases, and \(M_1\) and \(M_2\) are their molar masses.)
![**Calculate Pressure Using the Ideal Gas Law and the van der Waals Equation**
A 1.88-mol sample of argon gas is maintained in a 0.608-L container at 297 K. Calculate the pressure of the gas using both the ideal gas law and the van der Waals equation (van der Waals constants for Ar are \( a = 1.35 \, \text{L}^2\text{atm/mol}^2 \) and \( b = 3.22 \times 10^{-2} \, \text{L/mol} \)).
\[ P_{\text{ideal gas equation}} = \_\_\_\_ \, \text{atm} \]
\[ P_{\text{van der Waals}} = \_\_\_\_ \, \text{atm} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb8731af5-23ed-4fe5-8b12-d0cb96af1906%2Feb786cb1-dcae-46db-bb3e-b689524a0885%2Fen9ivl6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calculate Pressure Using the Ideal Gas Law and the van der Waals Equation**
A 1.88-mol sample of argon gas is maintained in a 0.608-L container at 297 K. Calculate the pressure of the gas using both the ideal gas law and the van der Waals equation (van der Waals constants for Ar are \( a = 1.35 \, \text{L}^2\text{atm/mol}^2 \) and \( b = 3.22 \times 10^{-2} \, \text{L/mol} \)).
\[ P_{\text{ideal gas equation}} = \_\_\_\_ \, \text{atm} \]
\[ P_{\text{van der Waals}} = \_\_\_\_ \, \text{atm} \]
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