Question

X g of Ag was dissolved in Hno3 and the solution was treated with excess of nacl when 2.87 g of agcl was precipitated. The value of x is..?

Answer 1

Ag + Cl = AgCl

Ag = 108

Cl = 35.5

AgCl = 143.5

143.5 g AgCl obtained from 108 g Ag

Therefore 2.87 g AgCl can be obtained from 108 x 2.87 / 143.5

= 108/50

= 2.17 g

x = 2.17 g of Ag

Answer 2

Ag+Cl⟶AgCl

Atomic mass of Ag=108 g

Atomic mass of Cl=35.5 g

Molecular mass of AgCl=143.5 g

143.5 g AgCl is obtained from 108 g Ag

2.87 g AgCl is obtained from =

143.5

108×2.87

=2.17 g