Question

1.25 GM of organic compound heated with nitric acid in the presence of AgNO3 in various tubes and 752 mg of AgBr is formed what is the weight (in mg) of Bromine present in 0.25 g of the same organic compound ?​

Answer 1

Step-by-step explanation:

According to Carius method,

% of Br=

108+80

80

×

Massoforganiccompound

MassofAgCl

×100

⟹% of Br=

188

80

×

0.186

0.320

×100

⟹% of Br=0.42×

0.186

0.320

×100

⟹% of Br=73.6%

Answer 2

$\huge \❥︎\color{red}\sf\: Question?$

1.25 GM of organic compound heated with nitric acid in the presence of AgNO3 in various tubes and 752 mg of AgBr is formed what is the weight (in mg) of Bromine present in 0.25 g of the same organic compound ?

$\huge\♡︎\color{gold}\sf\: Answer\♡︎$

$according \: \: to \: \: carius \: \: method$

$\% \: of \: br \: = \: \: \: \: \: \: \: \: \: \: \: \: 80 \ \\ \ \ \: \: : \: \: \: \: \: \: \: \: \: \: \:\: \ \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ---\ \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 108 +80$

$Mass \: \: \: of \: \: AgC \\ \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: ----- \: - (×100)</p><p> \\ Mass \: \: of \: \: organic \: \: compound</p><p>$

$\ = \% \: \: of \: \: br \: \: = \: \: 80 \: \: \: \: \: \: \: \: \: 0.320\\ \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: --- \times --- \: \: \times 100\: \: \: \\: \: \: \: \: \: \: \: \: \: ☞︎︎︎ \: \: \: ☞︎︎︎: \: \: \: \: \: \: 188 \: \: \: \: \: \: \: 0.186$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0.320\\ \% \: of \: br \: \: = 0.42 \: \times --- \times 100\\ \: \: \: \: \:☞︎︎ \: \: \: \: \: \: \: \:☞︎︎ \: \: \: \: \: 0.186$

% of BR => 73.6%