IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry - gaseous state(P Bahadur, pg no. 88, Q. No. 25)

dev_22oct

molecular wt of compd₁= 6.75 x 22.4 =151.2 g

there fore element's wt in compd₁= 96% x 151.2 =145.152 g......(1)

molecular wt of compd₂=9.56 x 22.4 =214.144 g

therefore element's wt in compd₂ =33.9% x 214.144 g =72.5 g....(2)

molecular wt of compd₃=10.08 x 22.4 = 225.792 g

therefore element's wt in compd₃ =96.4% x 225.792 =216.66 g......(3)

let n part of the element is present in compd₂

therefore (2) can be written as 72.5 = n x M .........(4) (M= atomic wt of element)

(1) can be written as 145.152 = 2n xM...........(5)

(3) can be written as 217.66 =3n xM............(6)

adding (4) (5) (6)

we get 6n x M =435.312 g

M = 435.312/6n = 72.5/n g