What is the molarity of cl− in each solution?

a. 0.210 m nacl
b. 0.160 m srcl2
c. 0.110m alcl3

The molarity of the Chloride ion, Cl¯ in each of the solutions are:

A. The molarity of Cl¯ in 0.210 M NaCl is 0.210 M.

B. The molarity of Cl¯ in 0.160 M SrCl₂ is 0.32 M

C. The molarity of Cl¯ in 0.110 M AlCl₃ is 0.33 M

A. Determination of the molarity of Cl¯ in 0.210 M NaCl

NaCl(aq) —> Na⁺(aq) + Cl¯(aq)

From the balanced equation above,

1 mole of NaCl produced 1 mole of Cl¯

Therefore,

0.210 M NaCl will also produce 0.210 M Cl¯

Thus, the molarity of Cl¯ in 0.210 M NaCl is 0.210 M  

B. Determination of the molarity of Cl¯ in 0.160 M SrCl₂

SrCl₂(aq) —> Sr²⁺(aq) + 2Cl¯(aq)

From the balanced equation above,

1 mole of SrCl₂ produced 2 moles of Cl¯.

Therefore,

0.160 M SrCl₂ will produce = 0.160 × 2 = 0.32 M Cl¯

Thus, the molarity of Cl¯ in 0.160 M SrCl₂ is 0.32 M

C. Determination of the molarity of Cl¯ in 0.110 M AlCl₃

AlCl₃(aq) —> Al³⁺(aq) + 3Cl¯(aq)

From the balanced equation above,

1 mole of AlCl₃ produced 3 moles of Cl¯.

Therefore,

0.110 M AlCl₃ will produce = 0.110 × 3 = 0.33 M Cl¯.

Thus, the molarity of Cl¯ in 0.110 M AlCl₃ is 0.33 M

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a. 0.210 M.

b. 0.32 M.

c. 0.33 M.

Explanation:

a. 0.210 M NaCl:

NaCl → Na⁺ + Cl⁻.

∴ [Cl⁻] = 0.210 M.

b. 0.160 M SrCl₂:

SrCl₂ → Sr²⁺ + 2Cl⁻.

∴ [Cl⁻] = 2 x (0.160) = 0.32 M.

c. 0.110M AlCl₃:

AlCl₃ → Al³⁺ + 3Cl⁻.

∴ [Cl⁻] = 3 x (0.110) = 0.33 M.

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