An organic compound contains 46.78 % of a hal | vedclass

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(C) Mass of $AgX = 2.21 \ g$.Mass of $X$ in $2.00 \ g$ of the compound $= \frac{46.78}{100} 	imes 2.00 \ g = 0.9356 \ g \approx 0.94 \ g$.Mass of $Ag

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$Br$

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(C) Mass of $AgX = 2.21 \ g$.Mass of $X$ in $2.00 \ g$ of the compound $= \frac{46.78}{100} 	imes 2.00 \ g = 0.9356 \ g \approx 0.94 \ g$.Mass of $Ag$ in $AgX = 	ext{Total mass of } AgX - 	ext{Mass of } X = 2.21 \ g - 0.94 \ g = 1.27 \ g$.Number of moles of $Ag = \frac{1.27 \ g}{108 \ g/mol} \approx 0.01176 \ mol$.Since the stoichiometry of $AgX$ is $1:1$,the number of moles of $X$ is also $0.01176 \ mol$.Atomic mass of $X = \frac{	ext{Mass of } X}{	ext{Moles of } X} = \frac{0.94 \ g}{0.01176 \ mol} \approx 79.93 \ g/mol$.This corresponds to the atomic mass of Bromine $(Br = 80)$.Therefore,the halogen $X$ is $Br$.

An organic compound contains $46.78 \%$ of a halogen $X$. When $2.00 \ g$ of this compound is heated with fuming $HNO_3$ in the presence of $AgNO_3$,$2.21 \ g$ of $AgX$ is formed. The halogen $X$ is:
[Given: atomic weight of $Ag = 108, F = 19, Cl = 35.5, Br = 80, I = 127$]

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An organic compound contains $46.78 \%$ of a halogen $X$. When $2.00 \ g$ of this compound is heated with fuming $HNO_3$ in the presence of $AgNO_3$,$2.21 \ g$ of $AgX$ is formed. The halogen $X$ is:[Given: atomic weight of $Ag = 108, F = 19, Cl = 35.5, Br = 80, I = 127$]