1.000 atm of Oxygen gas, placed in a container having apinhole opening in its si
ID: 688566 • Letter: 1
Question
1.000 atm of Oxygen gas, placed in a container having apinhole opening in its side. leaks from the container 2.14 timesfaster thatn 1.000 atm of an unknown gas placed in this sameapparatus. Which of the following species could be theunknown gas? A. CL2 B. SF6 C.Kr D.UF6 E. Xe Should I calcutate the molecular weight of each and usepricnipal of Grahams Law? Please show me how to make thisdetermination/ 1.000 atm of Oxygen gas, placed in a container having apinhole opening in its side. leaks from the container 2.14 timesfaster thatn 1.000 atm of an unknown gas placed in this sameapparatus. Which of the following species could be theunknown gas? A. CL2 B. SF6 C.Kr D.UF6 E. Xe Should I calcutate the molecular weight of each and usepricnipal of Grahams Law? Please show me how to make thisdetermination/ C.Kr D.UF6 E. Xe Should I calcutate the molecular weight of each and usepricnipal of Grahams Law? Please show me how to make thisdetermination/Explanation / Answer
According to Grahams Law of diffusion rate of diffusion, r 1 / M Where M is the Mol. wt. of the the gas For two cases r / r ' = ( M' / M ) r = rate of diffusion of unknown gas r' = rate of diffusion of O2 gas = 2.14 r M = Molar mass of unknown gas = ? M' = Molar mass of O2 = 2 * 16 = 32 g Plug the values we have r / r ' = ( M' / M ) r / 2.14 r = ( 32 / M ) ( 32 / M ) = 1 / 2.14 2 M = 146.5472 g Molar mass of SF6 is = 32 + 6 * 19 = 146 So the unknown gas is SF6.Related Questions
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