JEE MAIN - Chemistry (2015 (Offline) - No. 3)
In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The
percentage of bromine in the compound is: (at. Mass Ag = 108; Br = 80)
48
60
36
24
Explanation
Mass of substance $$=250$$ $$mg$$ $$=0.250$$ $$g$$
Mass of $$AgBr$$ $$=141$$ $$mg=0.141$$ $$g$$
$$1$$ mole of $$AgBr$$ $$=1$$ $$g$$ atom of $$Br$$
$$188$$ $$g$$ of $$AgBr$$ $$=80$$ $$g$$ of $$Br$$
$$188$$ $$g$$ of $$AgBr$$ contain bromine $$=80$$ $$g$$
$$0.141$$ $$g$$ of $$AgBr$$ contain bromine $$ = {{80} \over {188}} \times 0.141$$
This much amount of bromine present in $$0.250$$ $$g$$ of organic compound
$$\therefore$$ $$\% $$ of bromine $$ = {{80} \over {188}} \times {{0.414} \over {0.250}} \times 100 = 24\% $$
Mass of $$AgBr$$ $$=141$$ $$mg=0.141$$ $$g$$
$$1$$ mole of $$AgBr$$ $$=1$$ $$g$$ atom of $$Br$$
$$188$$ $$g$$ of $$AgBr$$ $$=80$$ $$g$$ of $$Br$$
$$188$$ $$g$$ of $$AgBr$$ contain bromine $$=80$$ $$g$$
$$0.141$$ $$g$$ of $$AgBr$$ contain bromine $$ = {{80} \over {188}} \times 0.141$$
This much amount of bromine present in $$0.250$$ $$g$$ of organic compound
$$\therefore$$ $$\% $$ of bromine $$ = {{80} \over {188}} \times {{0.414} \over {0.250}} \times 100 = 24\% $$
Comments (0)
