A gaseous mixture contains 1 part of H2 in every 65 parts by mass of a mixture of H2 and O2 gases. How many diffusion steps is needed such that the final gaseous mixture coming out should have 1 part of H2 in every 5 parts by mass of the mixture?
Ans.
Given, 1 in 65 parts is H2
Hence, parts of O2 = 65-1 = 64
Mass Ratio in parts can be written as 1:64
Therefore, Molar ratio of H2 to O2 = 1/2 : 64/32 [ Molar mass of H2=2 & Molar mass of O2=32]
= 1:4
Now, To have 1 part H2 in every 5 part of mixture
Part of 02 = 5-1 =4
Mass ratio in parts can be written as 1:4
Therefore molar ratio of H2 to O2 = 1/2 : 4/32
= 4:1
We know that, by comparing rate of effusion
R(H2)/R(O2)=nH2/nO2ｘ√M(O2)/√M(H2) [ r=rate , n=moles , m=molar mass ]
According to question,
4/1=1/4*(√32/2)^n
16=(√16)^n
Therefore, n=2
Therefore, no of diffusion steps required is 2.
