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 AgNO 3 [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)

Question

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 AgNO 3 [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) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

nc l in compound =nA g C l=(0.3849 g/(107.87+35.5)) g / mol ⇒ mass of chlorine =n Cl × 35.5=0.0953 gm ⇒ % wt of chlorine =(0.0953/0.5) × 100 =19.06 %

ncl​ in compound =nAgCl​=(107.87+35.5)0.3849g​g/mol ⇒ mass of chlorine =nCl​×35.5=0.0953gm ⇒% wt of chlorine =0.50.0953​×100 =19.06%

$n_{c l}$ in compound $= n_{A g Cl} = \frac{0.3849 g}{( 107.87 + 35.5 )} g / m o l$
$\Rightarrow$ mass of chlorine $= n_{Cl} \times 35.5 = 0.0953 g m$
$\Rightarrow %$ wt of chlorine $= \frac{0.0953}{0.5} \times 100$
$= 19.06%$