In carius method of estimation of halogens, 250 mg of an oraganic compound gave 141 mg of AgBr. The percentage of bromine in the compound is ( at mass: Ag= 108, Br=80)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 24
Weight of the given compound = 250 mg
Weight of the AgBr obtained from organic compound = 141 mg
Molecular weight of the bromine (X) = 80
Molecular weight of the silver bromide (AgBr) = 188
weight % of Br
$= \frac{Atomic weight of Br \times weight of AgBr \times 100}{Molecular mass of AgBr \times weight of compound}$
$= \frac{80 \times 141 \times 100}{188 \times 250}$
$= 24 %$

Suggest Corrections

Similar questions

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)

Q. In Carius method, of estimation of halogens,

250 m g

of an organic compound gave

141 m g

A g B r

. The percentage of bromine in the compound is:(Atomic mass

A g = 108; B r = 80

)

Q. In Carius method of estimation of halogens, 250

m g

of an organic compound gave 141

m g

A g B r

. The percentage of bromine in the compound is:
(Atomic mass of

A g

= 108, Br= 80)

Q. In Carius method of estimation of halogen 0.15 gram of an organic compound gave 0.12 gram of AgBr . calculate % of bromine in the compound. (Ag=108, Br=80)

Q. In the Carius method for the estimation of halogen, 0.18 g of an monobromo organic compound gave 0.12 g of

A g B r

. The percentage of bromine in the compound is:
(Atomic mass of Br = 80 u, Ag = 108 u)

Join BYJU'S Learning Program