7g of a sample of NaCl on treatment with
excess of silver nitrate gave 14.35g of AgCl.
NaCl in the sample is
1) 80% 2) 50% 3) 65.8% 4) 83.5%
Answers
Answered by
8
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...
Answered by
0
Answer:
Explanation:
NaCl+AgNO3 → AgCl+NaNO3
1 mole of NaCl gives 1 mole of AgCl
58.5g of NaCl gives 143.5g of AgCl
5.85g of NaCl gives 14.35g of AgCl
percentage of NaCl= 5.85÷7×100
=83.5 percent
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