Chapter 9, Problem 53A

Interpretation Introduction

Interpretation :

Mass of silver nitrate required to precipitate all the chloride ions as silver chloride must be calculated. The mass of silver chloride precipitate which will be obtained must also be calculated.

Concept Introduction :

One mole silver ion is required to precipitate one mole chloride ion. Amount of silver chloride from one mole silver cation will be one mole.

Answer to Problem 53A

Mass of silver nitrate required to precipitate all the chloride ions is 0.520 g.

The mass of silver chloride precipitate is 0.436 g.

Explanation of Solution

Mass of the chloride sample is 1.054 g.

Amount of Cl- is 10.3 % by mass $\frac{10.3\times 1.054}{100}\text{ g=}\frac{10.8562}{100}\text{ g =0}\text{.108 g}\text{.}$

The reaction is represented as follows:

  ${\text{AgNO}}_{\text{3}}{\text{(aq)+Cl}}^{\text{-}}\text{(aq)}\to \text{AgCl(s)}\text{.}$

Molar mass of AgNO₃, AgCl and Cl- are 170 g, 143.5 g and 35.5 g respectively.

As per balanced chemical equation,

35.5 g Cl- reacts with 170 g AgNO₃.

1 g Cl- reacts with $\frac{\text{170}}{\text{35}\text{.5}}\text{ g}$ AgNO₃.

Thus 0.108 g Cl- reacts with $\frac{\text{170}\times \text{0}\text{.108}}{\text{35}\text{.5}}\text{ g =}\frac{\text{18}\text{.36}}{\text{35}\text{.5}}\text{ g= 0}\text{.520 g}$ AgNO₃.

35.5 g Cl- produces 143.5 g AgCl.

1 g Cl- produces $\frac{\text{143}\text{.5}}{\text{35}\text{.5}}\text{ g}$ AgCl.

Thus 0.108 g Cl- produces $\frac{\text{143}\text{.5}\times \text{0}\text{.108}}{\text{35}\text{.5}}\text{ g =}\frac{\text{15}\text{.498}}{\text{35}\text{.5}}\text{ g= 0}\text{.436 g}$ AgCl.

Conclusion

Thus, the amount of silver nitrate is 0.520 g and amount of silver chloride is 0.436 g.