The rate of diffusion of a gas having molecular weight just double of N2 gas in 56 mLs-1 the rate of diffusion of N2 will be
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Molecular weight of Nitrogen gas is.
= 14 × 2 = 28
The molar mass of that other gas is :
2 × 28 = 56
By Graham's law of diffusion.
Rate A / Rate B = √(mmB / mmA)
Rate of Nitrogen =?
Rate of the gas = 56 ml /s
Doing substitution :
N / 56 = √56/28
N / 56 = √2
N = √2 × 56
N = 79.20
79. 20 ml / second
The rate of nitrogen is :
79.20 ml / s
Explanation:
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