Question

A gaseous mixture of O2 and X containing 20% of X diffused through a small hole in 134 seconds while pure O2 takes 124 seconds to diffuse through the same hole. find molecular weight of X.

Answer 1

According to Graham's law of diffusion, rate of diffusion of a gas is inversely proportional to the square root of the mass of the gas.

$\frac{Rate_{mixture}}{Rate_{O_{2}}} =\sqrt{\frac{Mass of O_{2}}{Mass of mixture}}$

Rate is inversely proportional to time.

$\frac{Rate_{mixture}}{RateO_{2}} = \frac{Time O_{2}}{Time_{mixture}}$

Plugging in the values to calculate X:

$\frac{124 s}{134 s}=\sqrt{\frac{32}{(Mass of mixture)g}}$

$(\frac{124}{134})^{2} =\frac{32}{(Mass of mixture)}$

0.856(Mass of mixture) = 32

Mass of mixture = 37.4 g

The mixture is 20% X and 80 % Oxygen

If we take total number of moles in the mixture as 1.

Moles of Oxygen = 0.80 mol

Mass of Oxygen = $0.80 mol * \frac{32 g}{1 mol}= 25.6 g$

Moles of X = 0.20 mol

Mass of X = 0.20 mol * Molecular weight of X

Mass of mixture = 0.8 * 32 + 0.2 * X

==> 0.8 * 32 + 0.2 * X = 21.9

25.6 + 0.2 X = 37.4

X = 58.9 g/mol

Answer 2

Chlidonia, Machi ur answer is Correct