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?

Question

Ellie Andrews · Accepted Answer

#37"g"#

Graham's law of effusion, which links the relative rates of gas effusion to their molar masses, can be applied.

#(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"#.

Wyatt Atkins · Answer

undefined To find the mass of Gas A, we can use Graham's law of effusion:

Rate of effusion A / Rate of effusion B = sqrt(Molar mass B / Molar mass A)

Given:
Rate of effusion A / Rate of effusion B = 0.68
Molar mass B = 17 g

Let's solve for the molar mass of Gas A:

0.68 = sqrt(17 / Molar mass A)

Squaring both sides:

0.4624 = 17 / Molar mass A

Multiplying both sides by Molar mass A:

Molar mass A = 17 / 0.4624

Molar mass A ≈ 36.73 g/mol