In Carius method of examination of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is _____ % (Nearest integer). (Given: Molar mass of Ag is 108 and Br is 80 g )

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
In Carius method of examination of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is _____ $\times 10^{- 1}$ % (Nearest integer).

(Given: Molar mass of Ag is 108 and Br is 80 g ${mol}^{- 1}$ ) [2025]

Ans.
(255)

$Molar mass of AgBr = (108 + 80) {g mol}^{- 1} = 188 {g mol}^{- 1}$

$Number of moles of AgBr (n_{AgBr}) = \frac{Mass of AgBr}{Molar mass of AgBr} = \frac{0.15}{188} mol$

$Moles of Br = Moles of AgBr = \frac{0.15}{188} mol$

$Mass of Br = Moles of Br \times Molar mass of Br = \frac{0.15}{188} \times 80 g$

$Mass % of Br = \frac{Mass of Br}{Mass of organic compound} \times 100$

$= \frac{\frac{0.15}{188} \times 80}{0.25} \times 100 = 25.53 % = 255.3 \times 10^{- 1} %$

Want Us to Email you About Special Offers & Updates?