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%
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B
9.64%
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C
8.45%
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D
7.23%
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Solution
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×moles ofAgCl+1×moles ofAgBr= Total moles of Ag in mixture
⇒1×x143.5+1×100−x188= Total moles of Ag in mixture
A mass balance of silver gives us: ⇒1×x143.5×108+1×100−x188×108=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, ⇒massofChlorine=number of moles of cholrine×molar mass of chlorine
=19.61143.5×35.5=4.85
Therefore, the amount of chlorine present in 100g mixture is 4.85g or percentage of chlorine present in mixture is 4.85%.