Question

in carius method of estimation of halogen 0.15 gram of an organic compound gives 0.12 gram of agbr find out the percentage of bromine in the compound​

Answer 1

Answer:

Percentage of Bromine = 34 %

Explanation:

We Have :-

0.15 gram of an organic compound gives 0.12 gram of AgBr

To Find :-

Percentage of bromine in the compound​

Solution :-

Mass of AgBr = 0.12 g

Molecular Mass of AgBr = 188 g/mol

Mass of Organic Compound = 0.15 g

80 g Br is present in 188 g AgBr

0.12 g AgBr contains Br = ( 80 / 188 ) * 0.12

                                       = 0.051 g

Percentage of Bromine = ( 0.051 / 0.15 ) * 100

                                       = 34 %

Percentage of Bromine = 34 %

Answer 2

Answer:

$\large\underline\textsf{33.3\% }$

$\tt\red{Given}\begin{cases}\tt\orange{Halogen\: =0.15\: gram } \\ \tt\orange{organic \:compound =0.12\:gram}\end{cases}$

$\small\underline{\underline{\texttt{\red{To\: Find }}}}$

$\small\sf{the \:percentage \:of \:bromine\: in \:the \:compound }$

$\huge\underline\textsf{Explantion:- }$

$\rightarrow\sf \red{mass \: of \: organic \: compound =0.15g}$

$\rightarrow\bf \blue{mass \: of \: agbr = 0.12g}$

$\rightarrow\sf\orange{ molecular \: mass \: of \: agbr = 188gmol {}^{ - 1} }$

$\rightarrow\bf \purple{188g \: agbr \: contain \: 80g \: bromine}$

$\rightarrow\sf \green{0.12 \: of \: agbr \: will \: contain \: \frac{\cancel8\cancel0}{1\cancel8\cancel8} \times \cancel{0.12}}$

$\rightarrow\bf\pink{0.05 \: of \: browmine}$

$\rightarrow\sf\purple{ percentage \: of \: browmine = \frac{\cancel{0.05}}{\cancel{0.15}} \times \cancel{100}}$

$\huge\underline{\rightarrow\bf 33.3\%}$