Question

A mixture of pure $AgCl$ and $AgBr$ is found to contain a total of 60.94% $Ag$ by mass. What is the mass % of chlorine in the mixture?
(Molar mass of Ag = 108 g/mol, Cl = 35.5 g/mol, Br = 80 g/mol)

• A. 4.85%
• B. 9.64%
• C. 8.45%
• D. 7.23%

Answer 1

The correct option is A 4.85%

Take 100 g of the mixture so it will contain 60.94 g of $Ag$.
Now, let us assume the mixture has $xg$ of $AgCl$ and $(100-x)g$ of $AgBr$.
Applying POAC on silver we get,
$1\times \text{moles of}AgCl+1\times \text{moles of}AgBr=\text{Total moles of Ag in mixture}$

$\Rightarrow 1\times \frac{x}{143.5}+1\times \frac{100-x}{188}=\text{Total moles of Ag in mixture}$

A mass balance of silver gives us:
$\Rightarrow 1\times \frac{x}{143.5}\times 108+1\times \frac{100-x}{188}\times 108=60.94$

From the above equation we get,
$x=19.61g$ of $AgCl$
Since the number of moles of $AgCl$ is equal to the number of moles of chlorine present in the mixture and the total mass of the mixture taken is $100g$,
$\Rightarrow massofChlorine=\text{number of moles of cholrine}\times \text{molar mass of chlorine}$

$=\frac{19.61}{143.5}\times 35.5=4.85$

Therefore, the amount of chlorine present in $100g$ mixture is $4.85g$ or percentage of chlorine present in mixture is 4.85%.