JEE MAIN - Chemistry (2021 - 27th July Morning Shift - No. 17)
An organic compound is subjected to chlorination to get compound A using 5.0 g of chlorine. When 0.5 g of compound A is reacted with AgNO3 [Carius Method], the percentage of chlorine in compound A is ____________ when it forms 0.3849 g of AgCl. (Round off to the Nearest Integer)
(Atomic masses of Ag and Cl are 107.87 and 35.5 respectively)
(Atomic masses of Ag and Cl are 107.87 and 35.5 respectively)
Answer
19
Explanation
Mass of organic compound = 0.5 gm.
Mass of formed AgCl = 0.3849 gm
% of Cl = $${{atomic\,mass\,of\,Cl \times mass\,formed\,AgCl} \over {molecular\,mass\,of\,AgCl \times mass\,of\,organic\,compound}} \times 100$$
$$ = {{35.5 \times 0.3849} \over {143.37 \times 0.5}} \times 100$$
= 19.06
$$ \approx $$ 19
Mass of formed AgCl = 0.3849 gm
% of Cl = $${{atomic\,mass\,of\,Cl \times mass\,formed\,AgCl} \over {molecular\,mass\,of\,AgCl \times mass\,of\,organic\,compound}} \times 100$$
$$ = {{35.5 \times 0.3849} \over {143.37 \times 0.5}} \times 100$$
= 19.06
$$ \approx $$ 19
Comments (0)
