1.

Rate of diffusion of NH_(3) is twice that of X. What is the molecular mass of X

Answer»

68
48
12
8

Solution :Rate of DIFFUSION, `r_(X)=((r_(NH_(3)))/(2))` (given )
Molecular MASS of `NH_(3),M_(NH_(3))=(N)+3(H)=14+3(1)=17`
Molecular mass of X,`M_(X)= ?`
By Graham's law of diffusion
`:. (r_(NH_(3)))/(r_(X))=sqrt((M_(X))/(M_(NH_(3))))` or ` 2= sqrt((M_(X))/(M_(NH_(3)))) `
`:. 4 xx M_(NH_(3))IMPLIES M_(X)=4 xx 17 = 68`


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