7g of a sample of NaCl on treatment withexcess of silver nitrate gave

1.	7g of a sample of NaCl on treatment withexcess of silver nitrate gave 14.35g of AgCl.NaCl in the sample is1) 80% 2) 50% 3) 65.8% 4) 83.5%
Answer» DEAR Student, ◆ Answer - (4) 83.5% ● Explaination - Reaction of sodium chloride with SILVER nitrtate is as follows - NaCl + AgNO3 --> AgCl + NaNO3 That is 1 mol of NaCl gives 1 mol of AgCl. Mass of NaCl PARTICIPATING in reaction is - W(NaCl) / M(NaCl) = W(AgCl) / M(AgCl) W(NaCl) / 58.45 = 14.35 / 143.5 W(NaCl) = 0.1 × 58.45 W(NaCl) = 5.845 g Percentage of NaCl in SAMPLE is - NaCl % = 5.845 / 7 × 100 NaCl % = 83.5 % Therefore, NaCl sample was 83.5 % pure. Hope that was useful...

7g of a sample of NaCl on treatment withexcess of silver nitrate gave 14.35g of AgCl.NaCl in the sample is1) 80% 2) 50% 3) 65.8% 4) 83.5%

Answer»

DEAR Student,

◆ Answer -

(4) 83.5%

● Explaination -

Reaction of sodium chloride with SILVER nitrtate is as follows -

NaCl + AgNO3 --> AgCl + NaNO3

That is 1 mol of NaCl gives 1 mol of AgCl.

Mass of NaCl PARTICIPATING in reaction is -

W(NaCl) / M(NaCl) = W(AgCl) / M(AgCl)

W(NaCl) / 58.45 = 14.35 / 143.5

W(NaCl) = 0.1 × 58.45

W(NaCl) = 5.845 g

Percentage of NaCl in SAMPLE is -

NaCl % = 5.845 / 7 × 100

NaCl % = 83.5 %

Therefore, NaCl sample was 83.5 % pure.

Hope that was useful...