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 x molar mass
=0.025 x
143.5 = 3.59 g

Heavy
water is used as

