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.

The % of KCl in the original sample would be as follows:

%

Given that,

Amount of sample of impure Potassium Chloride

Amount of Silver Nitrate precipitated

Now,

Molarity of KCl =

Minimum moles required for reaction

Final number of moles  

So,

Concentration = Number of moles * Molarity

Therefore,

% of KCl in the original Sample = Weight of Pure KCl/Weight of Impure KCl

×

%

Thus, % is the correct answer.

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97.78% KCl in the original sample

Explanation:

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