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
• B. 1.14 g
• C. 11.4 g
• D. 2 g

Answer 1

Answer: (A)

0.114 g

NaCl + $AgN{O}_3$ $\to$ AgCl + $NaN{O}_3$

(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 = $\frac{58.5\times 0.28}{143.5}$

= $\frac{16.38}{143.5}$

= 0.114 g of NaCl