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(D) The reaction of $4$-bromobenzyl alcohol with $Red \ P/Br_2$ replaces the $-OH$ group with $-Br$ to form $4$-bromobenzyl bromide $(R)$.The molecula

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$1.50$

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(D) The reaction of $4$-bromobenzyl alcohol with $Red \ P/Br_2$ replaces the $-OH$ group with $-Br$ to form $4$-bromobenzyl bromide $(R)$.The molecular formula of $R$ is $C_7H_6Br_2$.Molar mass of $R = (7 	imes 12) + (6 	imes 1) + (2 	imes 80) = 84 + 6 + 160 = 250 \ g/mol$.Moles of $R$ in $1.00 \ g = \frac{1.00 \ g}{250 \ g/mol} = 0.004 \ mol$.Each molecule of $R$ contains $2$ bromine atoms. Therefore,$1 \ mol$ of $R$ produces $2 \ mol$ of $AgBr$ in the Carius method.Moles of $AgBr$ formed $= 2 	imes 0.004 \ mol = 0.008 \ mol$.Molar mass of $AgBr = 108 + 80 = 188 \ g/mol$.Mass of $AgBr$ formed $= 0.008 \ mol 	imes 188 \ g/mol = 1.504 \ g \approx 1.50 \ g$.

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 . . . .
$\overline{[Given :}$ Atomic mass of $H = 1, C = 12, O = 16, P = 31, Br = 80, Ag = 108]$

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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 . . . .$\overline{[Given :}$ Atomic mass of $H = 1, C = 12, O = 16, P = 31, Br = 80, Ag = 108]$