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?

1 Answer
Jul 10, 2017

Answer:

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

Explanation:

We can use Graham's law of effusion to solve 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.