Question

In order to determine the composition of a mixture of halides containing $MB{r}_2$ and $Nal,14$ g mixture was dissolved in water. To this solution, $AgN{O}_3$ solution of certain molarity was added gradually. The mass of precipitate produced (in g) were measured and it was plotted against volume of $AgN{O}_3$ solution added (in ml). Its known that $AgI$ is precipitated first. Precipitation of $Br$ does not start until the already precipitating $I^{-}$ precipitates completely. Find out the value of $AB\times CD$ where:
$AB$ $=$ Atomic weight of metal $M$
$CD$$=$Mole percent of $Nal$ in original mixture.

• A. $4994$
• B. $2040$
• C. $3202$
• D. $3332$

Answer 1

The correct option is A $4994$

Mass of silver iodide precipitated $=9.4$ g.
Mass of silver bromide precipitated $=24.44-9.4=15.04$ g.
Mass of NaI $=9.4\times \frac{149.89}{234.77}=6$ g.
Mass of $MB{r}_2=14-6=8$ g
Mass of $MB{r}_2=15.04g\times \frac{M+159.8}{2\times 234.77}=8$ g
$M=89.9$ g/mol $=AB$
Moles of NaI $=\frac{6}{149.89}=0.04$
Moles of $MgB{r}_2=\frac{8}{89.9+159.8}=0.032$
Mole percent of Nal in original mixture $=CD=\frac{0.04}{0.04+0.032}=\times 100=55.55$%
The value of $ABCD$ is $55.55\times 89.9=4994$