The time taken for a certain volume of gas to diffuse through a small pore is 2 min under similar conditions an equal volume of oxygen took 5.65 min to pass .The molecular mass of gas is
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Use Graham’s Law of Effusion. [1] Rate is inversely proportional to the mass of the gas.
Hence rate1/rate2 = √M2/M1
But we are given the time taken for a given vol to diffuse through a hole. Now the faster the rate the shorter the time so instead of rate we must use the inverse of the time. So time is directly proportion to thde square root of the mass. Let 1 be O2 (M1 = 32)
2/5.65 = √M2/32
square both sides
M2 = [4/31.9]×32 = 4
[1]
Hence rate1/rate2 = √M2/M1
But we are given the time taken for a given vol to diffuse through a hole. Now the faster the rate the shorter the time so instead of rate we must use the inverse of the time. So time is directly proportion to thde square root of the mass. Let 1 be O2 (M1 = 32)
2/5.65 = √M2/32
square both sides
M2 = [4/31.9]×32 = 4
[1]
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