Question

50 POINTS!!!
Q.An impure sample of sodium chloride which weighed 1.2 g gave on treatment with excess of silver nitrate solution 2.4g of silver chloride as precipitate.calculate the percentage purity of the sample.
can somebody tell me how did we get the third last step
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Answer 1

Answer:

We have got the amount of pure NaCl which is 0.978g.

Total weight given is 1.20g

So to calculate the percentage, we have to first divide the purity amount by total amount and then multiply it with 100.

So we got the purity percentage 81.5%

Answer 2

Question :

An impure sample of sodium chloride which weighed 1.2 g gave on treatment with excess of silver nitrate solution 2.4g of silver chloride as precipitate. Calculate the percentage purity of the sample.

Solution :

$NaCl\: + \:AgNO_{3}$ → $AgCl\:+\:NaNO_{3}$

Here..

1 mole of NaCl precipitate into 1 mole of AgCl.

So,

Molar mass of NaCl = 23 + 35.5

=> 58.5 g

Molar mass of AgCl = 108 + 35.5

=> 143.5 g

Now..

143.5 g AgCl obtained from 58.5 g NaCl

So,

2.4 g of AgCl = 58.5/143.5

1 g of NaCl precipitate = 0.40767 × 2.4

=> 0.97841 g AgCl

So,

1.2 g of NaCl forms = 0.97841 AgCl

1 g NaCl forms = 0.97841/1.2

=> 0.81534 g of AgCl

0.81534 g of pure AgCl is formed. But we have to find the purity percentage.

So,

Percentage of purity = 0.81534 × 100

=> 81.534 %

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81.5% of pure sample of purity is obtained

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