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If two molecules of A and B having mass 100 kg and 64 kg and rate of diffusion of A is 12 xx 10^(-3), then what will be the rate of diffusion of B

Answer»

`15xx10^(-3)`
`64 xx 10^(-3)`
`5 xx 10^(-3)`
`46 xx 10^(-3)`

Solution :`m_(A)=((100)/(2))KG//` molecule `m_(B)=((64)/(2))kg//` molecule
Rate of diffusion, `r_(A) 12 xx 10^(-3)` and `r_(B) =?`
According to Graham's LAW of diffusion
`(r_(A))/(r_(B))=SQRT((d_(B))/(d_(A)))=sqrt((M_(B))/(M_(A)))`
Here, `M= ` molar mass and `d=` density, `m=` mass of one molecule.
`(r_(A))/(r_(B))=sqrt(((m_(B)xxN_(A)))/((m_(A)xxN_(A)))` where `N_(A)=` AVOGADROS number
`(r_(A))/(r_(B))=sqrt(((64//2))/((100//2)))=sqrt((64)/(100))=(8)/(10)=0.8`
or, `(12 xx 10^(-3))/(r_(B))=0.8 impliesr_(B)=15 xx 10^(-3)`


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