An organic compound contains 46.78 % of a hal | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$Br$

Vedclass · Answer

(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 $Ag = 108, F = 19, Cl = 35.5, Br = 80, I = 127$]

Similar Questions

Assign $A, B, C, D$ from the given type of reaction.
$Pb(NO_3)_2 + 2 KI \longrightarrow PbI_2 \downarrow + 2 KNO_3$

The number of moles of solute present in the solutions of $I$,$II$ and $III$ is respectively:
$I$. $500 \ mL$ of $0.2 \ M \ NaOH$
$II$. $200 \ mL$ of $0.1 \ N \ H_2SO_4$
$III$. $6 \ g$ of urea in $1 \ kg$ of water

Which of the following species has the highest mass?

Mole fraction of $O_2$ in $O_2$ and $O_3$ mixture is $0.25$. Hence percentage concentration $\left( \frac{w}{W} \% \right)$ of $O_2$ in the mixture is:

What is the approximate volume occupied by $22.4 \, L$ of water vapor at $STP$ when it is condensed into liquid water?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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 $Ag = 108, F = 19, Cl = 35.5, Br = 80, I = 127$]