a sample of unknown compound gas x is shown by analysis to contain sulphur and oxygen. The gas requires 28.3seconds to diffuse through a small aperture into a vacuum. An identical number of oxygen molecules pass through the same aperture in 20seconds. Determine the molecular mass of gas x (O=16, S=32)
Question
a sample of unknown compound gas x is shown by analysis to contain sulphur and oxygen. The gas requires 28.3seconds to diffuse through a small aperture into a vacuum. An identical number of oxygen molecules pass through the same aperture in 20seconds. Determine the molecular mass of gas x (O=16, S=32)
Solution
To solve this problem, we can use Graham's law of diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
The formula for Graham's law is:
Rate1/Rate2 = sqrt(M2/M1)
Where: Rate1 is the rate of diffusion of gas1 (in this case, oxygen) Rate2 is the rate of diffusion of gas2 (in this case, the unknown gas X) M1 is the molar mass of gas1 (oxygen) M2 is the molar mass of gas2 (the unknown gas X)
We know that the rate of diffusion of oxygen is faster than that of gas X, so we can set up the equation as follows:
20 sec/28.3 sec = sqrt(Mx/32)
Solving for Mx, we get:
Mx = 32 * (28.3/20)^2
Mx = 32 * 2
Mx = 64 g/mol
Therefore, the molecular mass of the unknown gas X is 64 g/mol. This suggests that the gas could be sulfur dioxide (SO2), as the molecular mass of SO2 is 64 g/mol (32 for sulfur and 32 for two oxygen atoms).
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