A gas whose molar mass you wish to know effuses through an opening at a rate one

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A gas whose molar mass you wish to know effuses through an opening at a rate one third as fast as that of helium gas. What is the molar mass of the unknown gas?

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Grahams law of effusion relates the rates of effusion of two gases to their molar masses The law states that Rate1 Rate2 Molar Mass2 Molar Mass1 Where Rate1 and Rate2 are the rates of effusion of gases 1 and 2 Molar Mass1 and Molar Mass2 are the molar masses of gases 1 and 2 In this case you are comparing an unknown gas lets call it Gas X to helium gas He Let RateX be the rate of effusion of Gas X and RateHe be the rate of effusion of helium Given that RateX is onethird as fast as RateHe we have RateX 13 RateHe Now we can set up the Grahams law equation RateX RateHe Molar MassHe Molar MassX Substitute the values 13 40026 gmol Molar MassX Now solve for Molar MassX Molar MassX 3 40026 gmol Molar MassX 3 40026 gmol2 Molar MassX 9 40026 gmol Molar MassX 360234 gmol So the molar mass ...View the full answer

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