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:

\[\ce{AgNO3 + KCl -> KNO3 + AgCl}\]

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

= `"Mass"/"Molar mass"`

= `4.25/170`

= 0.025 moles

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

= `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