Question

A mixture of $AgCl$ and $AgBr$ undergoes a loss in wt. by $5$% if it is exposed to excess chlorination. The % of $AgCl$ in the mixture.

• A. $21.12$%
• B. $78.88$%
• C. $5$%
• D. $95$%

Answer 1

The correct option is B $78.88$%

Let a and b be the AgCl and AgBr respectively in a mixture

According to the question

a+b=100 $Eq⟶1$

Molecular mass of AgCl=143 g

Molecular mass of AgBr=188 g

After Chlorination, AgBr changes into AgCl

188 g of AgBr changes into 143.5 g AgCl

if b gm AgBr changes into AgCl then AgCl is $\frac{143.5\times b}{188}$

Total AgCl after chlorination=a+$\frac{143.5\times b}{188}$

a+$\frac{143.5\times b}{188}$=95(given that 5% loss of mass occurs)$Eq⟶2$

Solving using equation 1 & 2

a=100-b $\left(FromEq1\right)$

Putting the value of a in $Eq2$

Equation 2 becomes

$100-b+\frac{143.5\times b}{188}=95b=21.12$

$a=100-21.12=78.88$

Therefore the AgCl % in th mixture is 78.88%