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

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Answer 1 · 2017-05-23T00:15:51

#37"g"#

We can use Graham's law of effusion , which relates the relative rates of effusion of gases in terms of their molar masses:

#(r_1)/(r_2) = sqrt((MM_2)/(MM_1)#

Where
#(r_1)/(r_2)# is the ratio of effusion of gas #1# to gas #2#, and
#MM_(1, 2)# is the molar mass of the gases #1# and #2#, respectively.

We're given that the ratio of effusion of gas A to gas B is #0.68#, and that #MM_B# is #17"g"/"mol"#, so

#0.68 = (sqrt17)/sqrt(MM_A)#

#MM_A = ((sqrt(17))/(0.68))^2 = 37"g"/"mol"#

The mass of the gas is thus #37"g"#.