Question

3.2 × 10^-3 moles of oxygen gas effuse from a container in 10.0 minutes. How many moles of CH4 gas could effuse from the same container in 10.0 minutes under the same conditions?​

Answer 1

Answer:

Explanation:

Graham’s law of diffusion or effusion of gases states that:

Rate2/Rate1=(M1/M2)^1/2, ——-(1) where

Rate1=(moles of gas1-O2)/t1, t1=10 min, being time taken for the given amount of gas1 to effuse.

Rate1=3.2 x 10^(-3) mol/10min

Rate2=(moles of gas2-CH4)/t2, t2=10, being time taken for the given amount of gas2 to effuse.

Rate2=y mol/10min; y is the required number of moles of CH4

M1=molar mass of gas1 (O2)=32g

M2=molar mass of gas2 (CH4)=16g

Substituting all known values in equation (1) gives:

y/[3.2x10^(-3)mol]=(32/16)^1/2

y=[(2)^1/2] x [3.2x10^(-3)]mol =4.525x10^(-3)mol=4.53x10^(-3) mol.

Thus the number of moles of CH4 effusing from the same container under the same conditions as the O2 is 0.00453 mole.