Question

Q32) 3.0L solution of BaCl2has a chloride ion concentration of 0.20M.
The barium ion concentration in this solution is
a. 0.067 M
c. 0.20 M
b. 0.10 M
d. 0.60 M

Answer 1

Answer:

BaCl2 = Ba2+ + 2Cl-

If [Cl-] = 0.2 mol/L then [Ba2+] 0.20/2 = 0.10mol/L. Now since we have 3-L solution, [Ba2+] = 3 x 0.1 = 0.3mole

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