From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm^(3)s^(-1). The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm^(3)s^(-1). Identify the gas. [Hint: Use Graham's law of diffusion : R_(1)//R_(2) = (M_(2)//M_(1))^(1//2), where R_(1), R_(2) are diffusion rates of gases 1 and 2, and M_(1) and M_(2) their respective molecular masses. The law is a simple consequence of kinetic theory.]

From a certain apparatus, the diffusion rate of hydrogen has an averag

1.	From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm^(3)s^(-1). The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm^(3)s^(-1). Identify the gas. [Hint: Use Graham's law of diffusion : R_(1)//R_(2) = (M_(2)//M_(1))^(1//2), where R_(1), R_(2) are diffusion rates of gases 1 and 2, and M_(1) and M_(2) their respective molecular masses. The law is a simple consequence of kinetic theory.]
Answer» Solution :Accoriding to Graham.s law of diffusion, `(r_1)/(r_2) sqrt((M_2)/(M_1))` Where, `r_(1)` = diffusion rate of HYDROGEN `= 28.7 cm^(3)s^(-1)` `r_(2)` = diffusion rate of unknown GAS `=7.2 cm^(3) s^(-1)` `M_(1)` = molecular mass of hydrogen = 2U `M_(2) = ?` `:. (28.7)/(7.2) = sqrt((M_2)/(2)) or M_(2) = ((28.7)/(7.2))^(2) xx 2` `= 31.78 ~~ 32`.