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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Let the molecular mass of x be m.
20% of this is :
0.2m
Molecular weight of oxygen gas = 32
Let the molecular weight of the mixture be n
The total molecular weight of the mixture is :
n
By Graham's law of diffusion :
134/124 = √(32)/(n)
1.081² = 32/(n)
1.1678 = 32/(n)
1.1678(n) = 32
1.1678n = 32
n = 32/1.1678
n = 27.40
20/100 × 27.40 = 5.48
The molecular mass of x = 5.48g
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