In Carius method of estimation of halogens, 355 mg of an organic compound gave 287 mg of AgCl. The percentage of chlorine in the compound is . (At. Mass: Ag =108 ; Cl =35.5 u )

Question

In Carius method of estimation of halogens, 355 mg of an organic compound gave 287 mg of AgCl. The percentage of chlorine in the compound is . (At. Mass: Ag =108 ; Cl =35.5 u ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

% of chlorine =(35.5 × text Mass of AgCl /143.5 × text Mass of organic compound ) × 100 =(35.5 × 0.287 × 100/143.5 × 0.355)=20 %

Q. In Carius method of estimation of halogens, $355 m g$ of an organic compound gave $287 m g$ of $A g Cl$ . The percentage of chlorine in the compound is______ .
(At. Mass: $A g = 108; Cl = 35.5 u$ )

$%$ of chlorine
$= \frac{35.5 \times Mass of A g Cl}{143.5 \times Mass of organic compound} \times 100$
$= \frac{35.5 \times 0.287 \times 100}{143.5 \times 0.355} = 20%$

Q. In Carius method of estimation of halogens, 355mg of an organic compound gave 287mg of AgCl. The percentage of chlorine in the compound is______ . (At. Mass: Ag=108;Cl=35.5u )

% of chlorine =143.5× Mass of organic compound 35.5× Mass of AgCl​×100 =143.5×0.35535.5×0.287×100​=20%

Q. In Carius method of estimation of halogens, $355 m g$ of an organic compound gave $287 m g$ of $A g Cl$ . The percentage of chlorine in the compound is______ .
(At. Mass: $A g = 108; Cl = 35.5 u$ )

$%$ of chlorine
$= \frac{35.5 \times Mass of A g Cl}{143.5 \times Mass of organic compound} \times 100$
$= \frac{35.5 \times 0.287 \times 100}{143.5 \times 0.355} = 20%$