Answer:
A. 0.160 M
B. 0.360 M
C. 0.450 M
Explanation:
A. Determination of molarity of Cl- in
0.160 M NaCl.
We shall write the dissociation equation for NaCl. This is illustrated below:
NaCl —> Na+ + Cl-
From the balanced equation above,
1 mole of NaCl produced 1 mole of Cl-
Therefore, 0.160 M NaCl will also produce 0.160 M Cl-
Therefore the molarity of Cl- is 0.160 M
B. Determination of molarity of Cl- in 0.180 M SrCl2
We shall write the dissociation equation for SrCl2. This is illustrated below:
SrCl2 —> Sr^2+ + 2Cl-
From the balanced equation above,
1 mole of SrCl2 produced 2 moles of Cl-
Therefore, 0.180 M SrCl2 will produce = 0.180 x 2 = 0.360 M Cl-
Therefore, the molarity of Cl- is 0.360 M
C. Determination of molarity of Cl- in 0.150 M AlCl3.
We shall write the dissociation equation for AlCl3. This is illustrated below:
AlCl3 —> Al^3+ + 3Cl-
From the balanced equation above,
1 mole of AlCl3 produced 3 moles of Cl-
Therefore, 0.150 M AlCl3 will produce = 0.150 x 3 = 0.450 M Cl-
Therefore, the molarity of Cl- is 0.450 M.
