Question

A mixture of $KBr$ and $NaBr$ weighing 0.560 g was treated with aq. $A{g}^{+}$ ion and all bromide ion was recovered as 0.970 g of pure $AgBr.$ What was the fraction by weight of $KBr$ in the sample?

• A. $0.1332$
• B. $0.2378$
• C. $0.4378$
• D. $0.3478$

Answer 1

The correct option is B $0.2378$

$KBr+NaBr\stackrel{A{g}^{+}}{\to }AgBr$

Let $x$ g be the weight of $KBr$ in the sample.
Then ($0.560-x$) g be the weight of $NaBr$.
Because no. of moles of $Br$ atoms are conserved in the reaction, applying POAC on $Br$ atom,
Moles of $Br$ in $KBr$ + moles of $Br$ in $NaBr$ = moles of $Br$ in $AgBr$
$\Rightarrow \frac{x}{119}+\frac{0.560-x}{103}=\frac{0.97}{188}$

$\Rightarrow x=0.1332$ g
Fraction of $KBr$ in mixture $=\frac{0.1332}{0.560}=0.2378$