Question

In the Carius method of estimation of halogens, 0.250g of an organic compound

gives 0.141g of AgBr. Calculate the percentage of bromine in the compound.

(at.wt. of Ag = 108, Br = 80)​

Answer 1

$\sf {\Large {\underline {ORGANIC\:CHEMISTRY}}}$ :

Molar Mass of AgBr = 108 + 80 = 188 g mol^(-1)

188 g AgBr contains Bromine = 80 g

=> 80 × 0.12 / 188 = 0.051 gm

Percentage of Bromine = ( 80/188 ) × ( Mass of AgBr/ Mass of Organic compound ) × 100

% of Bromine = 0.051 × 100 / 0.15 = 34 %

Answer 2

Answer:

mol^(-1)

188 g AgBr contains Bromine = 80 g

=> 80 × 0.12 / 188 = 0.051 gm

Percentage of Bromine = ( 80/188 ) × ( Mass of AgBr/ Mass of Organic compound ) × 100

% of Bromine = 0.051 × 100 / 0.15 = 34 %

Explanation:

answer is 34%