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Gas A effuses 0.68 times as fast as Gas B. If the molar mass of Gas B is 17 g/mol, what is the molar mass of Gas A?

Answer 1

Owen Ambrose

#"molar mass of A" = 37# #"g/mol"#

Graham's law of effusion can be applied to resolve this:

#(r_A)/(r_B) = sqrt((M_B)/(M_A))#

where

#(r_A)/(r_B)# is a number representing the ratio of effusion rates of gas A to gas B, in this case #0.68:1 = 0.68#

#M_A# and #M_2# are the molar masses of gases A and B, respectively.

We know the molar mass of gas B is #17# #"g/mol"#, so let's solve this for #M_A#:

#0.68 = sqrt(17color(white)(l)"g/mol")/(sqrt(M_B))#

#sqrt(M_B) = sqrt(17color(white)(l)"g/mol")/0.68#

#M_B = (sqrt(17color(white)(l)"g/mol")/0.68)^2 = color(red)(37# #color(red)("g/mol"#

rounded to #2# significant figures.

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Answer 2

Henry Antrim

The molar mass of Gas A is approximately 11.2 g/mol.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.