When a 0.4500 g sample of impure potassium chloride was dissolved in water and treated with an excess ofsilver nitrate, 0.8402 g of silver chloride was precipitated. Calculate the percentage KCl in the original sample.
Answers
Answer:
97.78% KCl in the original sample
Explanation:
The % of KCl in the original sample would be as follows:
[tex]97.78[/tex]%
Given that,
Amount of sample of impure Potassium Chloride [tex]= 0.4500 g[/tex]
Amount of Silver Nitrate precipitated [tex]= 0.8402g[/tex]
Now,
Molarity of KCl = [tex]74.5 g/mol[/tex]
Minimum moles required for reaction [tex]= 0.0059 mol[/tex]
Final number of moles [tex]= 0.0059 mol[/tex]
So,
Concentration = Number of moles * Molarity
[tex]= 0.0059 mol * 0.74.5 g/mol\\= 0.440g[/tex]
Therefore,
% of KCl in the original Sample = Weight of Pure KCl/Weight of Impure KCl
[tex]= 0.440/0.450[/tex] × [tex]100[/tex]
[tex]= 97.78[/tex]%
Thus, [tex]97.78[/tex]% is the correct answer.
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