Question

A precipitate of $AgCl$ and $AgBr$ has the mass $0.4066$ g. On heating a current of chlorine, the $AgBr$ is converted to $AgCl$ and the mixture loses $0.0725$ in mass. The $\%$ of mass in the original mixture is (as the nearest integer) :

Answer 1

Let the mass of $AgCl$ and $AgBr$ in the sample be $a$ and $b$ g respectively,

$\therefore a+b=0.4066...\left(i\right)$

On passing a current of ${Cl}_2$ through mixture, $AgBr$ changes to $AgCl$ according to equation :

$AgBr\stackrel{{Cl}_2}{\to }AgCl$; whereas $AgCl$ remains unaffected.

$\because$ $188$ g $AgBr$ changes to $143.5$ g $AgCl$.

$b$ g $AgBr$ changes to $\frac{143.5\times b}{188}$ g $AgCl$

Total mass of $AgCl$ after reaction $=$ Initial mass - mass loss

$a+\frac{143.5\times b}{188}=0.4066-0.0725...\left(ii\right)$

Solving equations (i) and (ii),

$a=0.1007g,b=0.3059g$

Mass of $Cl$ in $0.1007$ g $AgCl$ $=35.5\times 0.1007143.5=0.025$

% of $Cl$ in mixture $=(0.025\times 100)/0.4066=6.15$%

So, answer is $6$.