Consider the following reaction. On estimation of bromine in 1.00 g of R using Carius method, the amount of AgBr formed (in g ) is [Given: Atomic mass of H =1, C =12, O =16, P =31, Br =80, Ag =108 ]

Question

Consider the following reaction. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/9137a8dbab7ab48eac2cf7c8b966aef1-.png" /> On estimation of bromine in 1.00 g of R using Carius method, the amount of AgBr formed (in g ) is [Given: Atomic mass of H =1, C =12, O =16, P =31, Br =80, Ag =108 ] (A)  (B)  (C)  (D)

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/ef8d907a2bbd3cab082f833398de8474-.png" /> operatornamelg~R arrow (1/250) moles No. of Br Atoms arrow (2/250) moles Moles of AgBr arrow (2/250) moles Mass of AgBr =(2/250) ×(108+80)=1.504

Q. Consider the following reaction.

On estimation of bromine in $1.00 g$ of $R$ using Carius method, the amount of $A g B r$ formed (in $g$ ) is __
[Given : Atomic mass of $H = 1, C = 12, O = 16, P = 31, B r = 80, A g = 108$ ]

$lg R \to \frac{1}{250}$ moles
No. of Br Atoms $\to \frac{2}{250}$ moles
Moles of $A g B r \to \frac{2}{250}$ moles
Mass of $A g B r = \frac{2}{250} \times (108 + 80) = 1.504$

Q. Consider the following reaction. On estimation of bromine in 1.00g of R using Carius method, the amount of AgBr formed (in g ) is __ [Given : Atomic mass of H=1,C=12,O=16,P=31,Br=80,Ag=108 ]

lg R→2501​ moles No. of Br Atoms →2502​ moles Moles of AgBr→2502​ moles Mass of AgBr=2502​×(108+80)=1.504

Q. Consider the following reaction.

On estimation of bromine in $1.00 g$ of $R$ using Carius method, the amount of $A g B r$ formed (in $g$ ) is __
[Given : Atomic mass of $H = 1, C = 12, O = 16, P = 31, B r = 80, A g = 108$ ]

$lg R \to \frac{1}{250}$ moles
No. of Br Atoms $\to \frac{2}{250}$ moles
Moles of $A g B r \to \frac{2}{250}$ moles
Mass of $A g B r = \frac{2}{250} \times (108 + 80) = 1.504$