In Carius method of estimation of halogen. 0.45 g of an organic compound gave 0.36 g of AgBr. Find out the percentage of bromine in the compound. (Molar masses: AgBr =188 g mol -1: Br =80 g mol -1 )

Question

In Carius method of estimation of halogen. 0.45 g of an organic compound gave 0.36 g of AgBr. Find out the percentage of bromine in the compound. (Molar masses: AgBr =188 g mol -1: Br =80 g mol -1 ) (A) 34.04 % (B) 40.04 % (C) 36.03 % (D) 38.04 %

Tardigrade Admin · Accepted Answer

Mass of organic compound =0.45 gm Mass of AgBr obtained =0.36 gm ∴ text Moles of AgBr =(0.36/188) ∴ Mass of Bromine =(0.36/188) × 80=0.1532 gm ∴ % Br in compound =(0.1532/0.45) × 100=34.04 %

Q. In Carius method of estimation of halogen. $0.45 g$ of an organic compound gave $0.36 g$ of $A g B r$ . Find out the percentage of bromine in the compound. (Molar masses : $A g B r = 188 g m o l^{- 1}$ : $B r = 80 g m o l^{- 1}$ )

$34.04%$

$40.04%$

$36.03%$

$38.04%$

Mass of organic compound $= 0.45 g m$
Mass of $A g B r$ obtained $= 0.36 g m$
$∴ Moles of A g B r = \frac{0.36}{188}$
$∴$ Mass of Bromine
$= \frac{0.36}{188} \times 80 = 0.1532 g m$
$∴ % B r$ in compound
$= \frac{0.1532}{0.45} \times 100 = 34.04%$

Q. In Carius method of estimation of halogen. 0.45g of an organic compound gave 0.36g of AgBr. Find out the percentage of bromine in the compound. (Molar masses : AgBr=188gmol−1 : Br=80gmol−1 )

34.04%

40.04%

36.03%

38.04%