Question

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.

Answer 1

Answer:

97.78% KCl in the original sample

Explanation:

Answer 2

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

$97.78$%

Given that,

Amount of sample of impure Potassium Chloride $= 0.4500 g$

Amount of Silver Nitrate precipitated $= 0.8402g$

Now,

Molarity of KCl = $74.5 g/mol$

Minimum moles required for reaction $= 0.0059 mol$

Final number of moles  $= 0.0059 mol$

So,

Concentration = Number of moles * Molarity

$= 0.0059 mol * 0.74.5 g/mol\\= 0.440g$

Therefore,

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

$= 0.440/0.450$ × $100$

$= 97.78$%

Thus, $97.78$% is the correct answer.

Learn more about "Potassium Chloride" here:

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