Question

1.11 g of the chloride of a metal dissolved in water were treated with an excess of silver nitrate solution .the weight of the precipitated silver chloride after washing and drying was found to be 2.87 g .calculate the equivalent weight of the metal

Answer 1

Answer:

Explanation: AgNO3 + Mcl - AgCl + M(NO3)x

Moles of AgCl = 2.87÷143 = 0.02 mole