Question

What is the mass of precipitate formed when 50 ml of 8.5% solution of AgNO₃ is mixed with 100 ml of 1.865% potassium chloride solution?

Options

• 3.59 g

• 7 g

• 14 g

• 28 g

Answer 1

3.59 g

Explanation:

$$\mathrm{AgNO}{\phantom{A}}_3+\mathrm{KCl}⟶\mathrm{KNO}{\phantom{A}}_3+\mathrm{AgCl}$$

50 mL of 8.5% solution contains 4.25 g of AgNO₃

No. of moles of AgNO₃ present in 50 mL of 8.5% AgNO₃ solution

= ${\displaystyle \frac{\text{Mass}}{\text{Molar mass}}}$

= ${\displaystyle \frac{4.25}{170}}$

= 0.025 moles

Similarly, No of moles of KCl present in 100 mL of 1.865% KCl solution

= ${\displaystyle \frac{1.865}{74.5}}$

= 0.025 moles

So total amount of AgCl formed is 0.025 moles (based on the stoichiometry).

Amount of AgCl present in 0.025 moles of AgCl

= No. of moles × molar mass

= 0.025 × 143.5

= 3.59 g