A gaseous mixture of O2 and x containing 20% of x, diffused through a small hole in 134 sec while pure O2 takes 124 secs to diffuse through the same hole. Find molecular weight of x.
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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.
Rate is inversely proportional to time.
Plugging in the values to calculate X
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 =
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
Rate is inversely proportional to time.
Plugging in the values to calculate X
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 =
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
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