Behavior Of Gases: Molar Mass Of a Vapor

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Problem #1: If equal amounts of helium and argon ar placed in a poriferous container and allowed to escape, which gas entrust escape quicker and how much faster?
dissolvent:
stupefy place1 = He = x
slump rate2 = Ar = 1
The molecular(a) saddle of He = 4.00
The molecular pack of Ar = 39.95
Grahams Law is:
r1 over r2 = ?MM2 over ?MM1
Substituting, we have:
x / 1 = ?(39.95 / 4.00)
x = 3.16 times as fast.

Video: Solution to a Grahams Law Problem

Problem #2: What is the molecular metric clog unit of a gas which diffuses 1/50 as fast as hydrogen?
Solution:
Set rate1 = opposite gas = 1
Set rate2 = H2 = 50
The molecular weight of H2 = 2.02
The molecular weight of the other gas = x.
By Grahams Law (see the answer to question #1), we have:
1 / 50 = ?(2.02 / x)
x = 5050 g/mol

Video: Solution to a Grahams Law Problem

Problem #3: Two porous containers are filled with hydrogen and neon respectively. Under identical conditions, 2/3 of the hydrogen escapes in 6 hours. How long will it take for half the neon to escape?
Solution:
Set rate1 = H2 = x
Set rate2 = Ne = 1
The molecular weight of H2 = 2.02
The molecular weight of Ne = 20.18
By Grahams Law:
x / 1 = ?(20.18 / 2.02)
x = 3.



16
Since the H2 escapes 3.16 times as fast as Ne, this calculation determines the amount of Ne leaving in 6 hours:
0.67 / 3.16 = 0.211
Calculate the time needed for half the Ne to escape, keen that 0.211 escapes in 6 hours:
0.211 / 6 = 0.50 / x
x = 14.2 hours

Problem #4: If the density of hydrogen is 0.090 g/L and its rate of diffusion is 5.93 times that of chlorine, what is the density of chlorine?
Solution:
Set rate1 = H2 = 5.93
Set rate2 = Cl2 = 1
The molecular weight of H2 = 2.02
The molecular weight of Cl2 = x.
By Grahams Law:
5.93 / 1 = ?(x / 2.02)
x = 71.03 g/mol
Determine gas density using the molar book of account:
71.03 g / 22.414 L =...If you want to get a near essay, order it on our website: Orderessay



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