Question

What is the mass of precipitate formed when 50 ml of 8.5% solution of Ag NO₃ is mixed with 100 ml of 1.865% potassium chloride solution? 

(a) 3.59 g 

(b) 7 g 

(c) 14 g 

(d) 28 g

Answer 1

(a) 3.59 g

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 loo 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