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Question

A solution of common salt when added to silver nitrate solution yields a precipitate of silver chloride(0.28 g). Find the mass of sodium chloride in the solution.


A

0.114 g

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B

1.14 g

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C

11.4 g

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D

2 g

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Solution

The correct option is A

0.114 g


NaCl + AgNO3 AgCl + NaNO3

(23+35.5) (108+35.5) (23+14+48)

58.5 143.5 85

Since, 143.5 g of silver chloride is formed from 58.5 g of NaCl,

0.28 g of silver chloride is formed = 58.5×0.28143.5

= 16.38143.5

= 0.114 g of NaCl


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