Question
An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (III) chloride. The number of moles of AgCl precipitate would be
-
0.001
-
0.002
-
0.003
-
0.01
Solution
A.
0.001
The formula of dichlorotetraqua chromium (III) chloride is [Cr(H2O)Cl2]Cl.
On ionisation it generates only one Cl- ion.
[Cr(H2O)4Cl2]Cl --> [Cr(H2O)Cl2]+ +Cl-
Initial 100x0.01 0 0
after ionisation
0 1 mol 1 mol
One mole of Cl- ions reacts with only 1 mole of AgNO3 molecules to produce 1 mole of AgCl.
therefore, 1 mmol or 1 x 10-3 mole reacts with AgNO3 to give AgCl.
= 1 x 1 x 10-3/1 = 0.001 mol AgCl
0 1 mol 1 mol
One mole of Cl- ions reacts with only 1 mole of AgNO3 molecules to produce 1 mole of AgCl.
therefore, 1 mmol or 1 x 10-3 mole reacts with AgNO3 to give AgCl.
= 1 x 1 x 10-3/1 = 0.001 mol AgCl
