Question

In Carius method a known mass of the organic compound is heated with excess of fuming $HN{O}_3$ and a few crystals of $AgN{O}_3$ in a sealed tube called Carius tube. $C$ an $H$ are oxidised to $C{O}_2$ and $H_{2}O$ respectively and the halogens are converted into silver halides. The ppt. of silver halide is filtered, washed, dried and weighted. The percentage of halogen can be calculated from the mass of silver halide formed
Percentage of $Cl=\frac{35.5}{143.5}=\frac{\text{ mass of AgCl formed }}{\text{ mass of sub taken }}\times 100$
Percentage of $Br=\frac{80}{188}=\frac{\text{ mass of AgCl formed }}{\text{ mass of sub taken }}\times 100$
Percentage of $I=\frac{127}{235}=\frac{\text{ mass of AgCl formed }}{\text{ mass of sub taken }}\times 100$
When $0.35g$ of an organic compound is heated with $HN{O}_3$ and $AgN{O}_3$ in a Carius tube, it gives $0.70g$ of silver chloride. The percentage of chloride in the compound is

• A. $54.8\%$
• B. $49.47\%$
• C. $34.6\%$
• D. $25.85\%$

Answer 1

Answer: (B)

$\%$ of chlorine $=\frac{35.5}{143.5}\times \frac{\text{ wt.of AgCl }}{\text{ wt.of organic compound }}\times 100$
$=\frac{35.5}{143.5}\times \frac{0.70}{0.35}\times 100=49.47\%$