In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. The percentage of bromine in the compound is:

43%

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

34.04%

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

34%

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

None of these

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

Open in App

Solution

The correct option is B
34.04%

Molar mass of AgBr = 108 + 80 = 188 g mol-1

188 g AgBr contains 80 g bromine

0.12 g AgBr contains= 80 X 0.12/188 g bromine

Percentage of bromine=80 X0.12 X100/188 X 0.15

= 34.04%

Suggest Corrections

Similar questions

Q. In the Carius method for the estimation of halogen, 0.15 g of an monobromo organic compound gave 0.12 g of AgBr. The percentage of bromine in the compound is:
(Atomic mass of Br = 80 u, Ag = 108 u)

Q. In Carius method of estimation of halogen,

1

g of an organic compound gave

0.094

g of

AgBr

. The percentage of bromine in the compound is :

Q. 0.15 g of an organic compound gave 0.12g of AgBr by Carius method. The % of bromine in the Compound is(Molar mass of AgBr=188)

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

In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. The percentage of bromine in the compound is:

The correct option is B 34.04% Molar mass of AgBr = 108 + 80 = 188 g mol-1 188 g AgBr contains 80 g bromine 0.12 g AgBr contains= 80 X 0.12/188 g bromine Percentage of bromine=80 X0.12 X100/188 X 0.15 = 34.04%

The correct option is B
34.04%

Molar mass of AgBr = 108 + 80 = 188 g mol-1

188 g AgBr contains 80 g bromine

0.12 g AgBr contains= 80 X 0.12/188 g bromine

Percentage of bromine=80 X0.12 X100/188 X 0.15

= 34.04%