X gm of Ag was dissolved in HNO3 and thesolution was treated with exce

1.	X gm of Ag was dissolved in HNO3 and thesolution was treated with excess of NaClWhen 2.87 gm of AgCl was precipitated thevalue of x is(A) 1.08 gm(C) 2.70 gm(B) 2.16 gm(D) 1.62 gm
Answer» 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 Agoption b

X gm of Ag was dissolved in HNO3 and thesolution was treated with excess of NaClWhen 2.87 gm of AgCl was precipitated thevalue of x is(A) 1.08 gm(C) 2.70 gm(B) 2.16 gm(D) 1.62 gm

Answer»

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 Agoption b