Under certain conditions of temperature and pressure this gas effuses through a pinhole at a rate of 16.60 mL in 10
minutes. At the same temperature and pressure argon effuses through the same pinhole at
a rate of 15.0 mL in 5 minutes. Determine the molecular mass of gas X.
minutes. At the same temperature and pressure argon effuses through the same pinhole at
a rate of 15.0 mL in 5 minutes. Determine the molecular mass of gas X.
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r1 / r2 = √(MM2 / MM1)
Let r2 be the effusion rate for X: 16.60 mL / 10 min = 1.660 mL/min
Let r1 be the effusion rate for Ar: 15.0 mL / 5 min = 3.0 mL/min
3.0 / 1.66 = √(x / 39.948 g/mol)
3.266 = x / 39.948 g/mol
x = 130.5 g/mol
Let r2 be the effusion rate for X: 16.60 mL / 10 min = 1.660 mL/min
Let r1 be the effusion rate for Ar: 15.0 mL / 5 min = 3.0 mL/min
3.0 / 1.66 = √(x / 39.948 g/mol)
3.266 = x / 39.948 g/mol
x = 130.5 g/mol
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The higher the molecular weight of a gas, the lower the effusion or leaking rate through a small hole such as in a hole in a car tire.
the fomula for this is Leak rate gas A over leak rate gas B = square root molec wt of gas B
over sq. root molec wt gas A
so fill in the numbers I will call the unknown gas, A and the Argon will be gas B
and realize that if l5 ml Ar leaks in 5 minutes, then 30 ml will leak in 10 minutes.
l6.60 ml gas A over 30 ml Argon = square root molec wt. Argon over
square root molecular wt gas A
set up this math and cross multiply
the fomula for this is Leak rate gas A over leak rate gas B = square root molec wt of gas B
over sq. root molec wt gas A
so fill in the numbers I will call the unknown gas, A and the Argon will be gas B
and realize that if l5 ml Ar leaks in 5 minutes, then 30 ml will leak in 10 minutes.
l6.60 ml gas A over 30 ml Argon = square root molec wt. Argon over
square root molecular wt gas A
set up this math and cross multiply