Explanation:
White precipitate of silver chloride get dissolves in excess ammonia to formation of complex between silver ions, chloride ions and ammonia molecules.
The chemical reaction is given as:
![AgCl(s)+2NH_3(aq)\rightarrow Ag[(NH_3)_2]^+Cl^-(aq)](https://tex.z-dn.net/?f=AgCl%28s%29%2B2NH_3%28aq%29%5Crightarrow%20Ag%5B%28NH_3%29_2%5D%5E%2BCl%5E-%28aq%29)
When 1 mole of silver chloride is added to 2 mole of an aqueous ammonia it form coordination complex of diaaminesilver(I) chloride.
The answer should be D all of the above
Answer:
The answer to your question is: 0.1 M
Explanation:
data
Volume of AgNO3 = 20.00 ml
1000 ml -------------- 1 l
20 ml --------------- x
x = 20x 1 /1000 = 0.02
AgCl = 0.2867 g
MW of AgCl = 35.45 + 107.9 = 143.35 g
143.35 g -------------- 1 mol
0.2867 g ------------- x
x = 0.2867 x 1 / 143.35 = 0.002 moles of AgCl
From the balance reaction we see that the proportion of AgNO3 to AgCl is 1:1, then
1 mol of AgNO3 -------------------- 1 mol of AgCL
x --------------------- 0.002 moles of AgCl
x = 0.002 moles of AgNO3
This moles of AgNO3 are in 20 ml or 0.02 liters
So, Molarity = # moles/liter
Molarity = 0.002 moles/ 0.02 = 0.1 M
