A gaseous mixture contains oxygen and another unknown gas in the molar ratio 4:1 diffuses through a porous plug in 245 seconds. Under simillar conditions same volume of oxygen takes 220 seconds to diffuse. Find the molecular mass of the unknown gas.

Here we will use the Graham's law of diffusion. According to the Graham's law of diffusion

r_{1} / r_{2} = (d_{2} / d_{1}) ^{1/2}

where r_{1} and r_{2} are the rate of diffusion of the two gases, d_{1} and d_{2} are their densities. The rate of diffusion or effusion of any gas is defined as

= (volume of the gas effused or diffused ) / Time taken

We are given that the mixture of oxygen and the unknown gas diffuse in 245 seconds. The same volume of only oxygen gas diffuses in 220 seconds. Let x grams be the molar mass of the unknown gas. So the molar mass of mixture will be mass of oxygen gas + x gmol^{-1} = (32 + x) gmol^{-1}.

Thus we will have

r_{mixture} / r_{oxygen} = (d_{oxygen} / d_{mixture}) ^{1/2}

We are given that the volume of oxygen gas and the mixture that diffuse are the same, therefore the above equation reduces to

t_{oxygen} / t_{mixture} = (M_{oxygen} / M_{mixture}) ^{1/2}

220 / 245 = [ 32 / (x + 32)] ^{1/2}

Solving for x, we get

x = 7.685 gmol^{-1}

Therefore the molar mass of the gas is 7.685 gmol^{-1}

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