Question

A mixture of two gases was allowed to effuse from a container. one of the gases escaped from the container 1.43 times as fast as the other one. the two gases could have been:

Answer 1

Answer:

C. Cl2 and SF6

Explanation:

• According to the Graham's law of diffusion the rate of effusion or diffusion of a fixed mass of a gas is inversely proportional to the square root of its density or mass.

• For two gases; R1/R2 = √mm2/√mm1 ; where R1 and R2 are the rates of diffusion of the two gases  respectively, while mm1 and mm2 are the molecular masses respectively.

Therefore, in this case we need two gases that have a ratio of their rates of diffusion as 1.43.

Thus; For Cl2 and SF6

Molecular mass of Cl2 is 70.906 g/mol, while that of SF6 is 146.06 g/mol

Therefore, the ratio of their rates of diffusion will be;

√(MM of SF6)/√(MM of Cl2) = √ 146.06/√70.906

                                              = 1.4352

                                              = 1.43

Therefore; the two gases required are Cl2 and SF6