200cm3 of a diffused through a porous plug in 5s under the same condition 450cm3of oxygen diffused in 5s. Calculate the relative molecular mass of the gas

Answer 1

First, we need to use Graham's law of diffusion to find the ratio of the rates of diffusion of the two gases:

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

We know that the rate of oxygen (gas B) is 450 cm3 in 5 seconds, or 90 cm3/s.

Let x be the relative molecular mass of the other gas (gas A), and let its rate of diffusion be y cm3/s.

Then, we can write:

y / 90 = sqrt (32 / x)

where 32 is the molar mass of oxygen (in g/mol).

Next, we can use the given data that 200 cm3 of gas A diffused in 5 seconds to solve for y:

y / 90 = 200 / 450

y = 80 cm3/s

Substituting into the previous equation, we can solve for x:

80 / 90 = sqrt (32 / x)

x = (32 * 90^2) / 80^2

x = 36

Therefore, the relative molecular mass of gas A is 36.