91影视

To determine the molar mass of each molecule, we need to add the molar mass of hydrogen and the molar mass of chlorine isotope. The molar masses for each isotope should be provided either in the question or in a periodic table, but for this exercise we will assume they are as follows: - \(^{1}\mathrm{H}\): 1 g/mol - \(^{2}\mathrm{H}\): 2 g/mol - \(^{35}\mathrm{Cl}\): 35 g/mol - \(^{37}\mathrm{Cl}\): 37 g/mol Using these values, we can now determine the molar masses for each of the four molecules: 1. \(^{1}\mathrm{H}^{35}\mathrm{Cl}\): 1 + 35 = 36 g/mol 2. \(^{1}\mathrm{H}^{37}\mathrm{Cl}\): 1 + 37 = 38 g/mol 3. \(^{2}\mathrm{H}^{35}\mathrm{Cl}\): 2 + 35 = 37 g/mol 4. \(^{2}\mathrm{H}^{37}\mathrm{Cl}\): 2 + 37 = 39 g/mol

Graham's Law states that the rate of effusion is inversely proportional to the square root of the molar mass. Mathematically, this means that: Rate of effusion \(鈭� \dfrac{1}{\sqrt{Molar\ Mass}}\) Now, we can compare the rates of effusion for the four molecules based on their molar masses: 1. \(^{1}\mathrm{H}^{35}\mathrm{Cl}\): \(\dfrac{1}{\sqrt{36}}\) = \(\dfrac{1}{6}\) 2. \(^{1}\mathrm{H}^{37}\mathrm{Cl}\): \(\dfrac{1}{\sqrt{38}}\) 3. \(^{2}\mathrm{H}^{35}\mathrm{Cl}\): \(\dfrac{1}{\sqrt{37}}\) 4. \(^{2}\mathrm{H}^{37}\mathrm{Cl}\): \(\dfrac{1}{\sqrt{39}}\) It should be noted that the values don't need to be simplified, as we're only interested in comparing the rates. Since larger molar mass results in a smaller rate of effusion and smaller molar mass results in a larger rate, we can now put the molecules in order of increasing rate of effusion: 1. \(^{2}\mathrm{H}^{37}\mathrm{Cl}\) - the lowest rate (highest molar mass) 2. \(^{1}\mathrm{H}^{37}\mathrm{Cl}\) 3. \(^{2}\mathrm{H}^{35}\mathrm{Cl}\) 4. \(^{1}\mathrm{H}^{35}\mathrm{Cl}\) - the highest rate (lowest molar mass)