What is the molarity in Cl- in each solution? Part A) 0.160 M NaCl
Part B) 0.180 M SrCl2
Part C) 0.150 M AlCl3

A) The molarity in  Cl⁻  0.160 M NaCl = 0.160 M

B) The molarity in Cl⁻  0.180 M SrCl₂ = 0.360 M

C) The molarity in Cl⁻ 0.150 M AlCl₃ = 0.450 M

A) Determine the molarity in  Cl⁻  0.160 M NaCl

first step : write out the balanced dissociation reaction equation for NaCl

NaCl > Na⁺ + Cl⁻

From the balanced dissociation equation : 1 mole of NaCl = 1 mole of Cl⁻

∴ 0.160 M of NaCl = 0.160 M of Cl⁻

B) Determine The molarity in Cl⁻  0.180 M SrCl₂

First step : write out the balanced dissociation reaction equation for SrCl₂

SrCl₂  ---> Sr²⁺  +  2Cl⁻

From the equation ; 1 mole of SrCl₂ = 2 moles of Cl⁻

∴ 0.180 M of  SrCl₂ = 0.180 * 2 = 0.360 M of Cl⁻

C) Determine the molarity in Cl⁻ 0.150 M AlCl₃

First step : write out the balanced dissociation reaction equation for AlCl₃

AlCl₃ --->  Al³⁺ + 3Cl⁻

From the balanced dissociation equation

0.150 M AlCl₃ will produce = 0.150 * 3 = 0.450 M  of Cl⁻

Hence we can conclude that A) The molarity in  Cl⁻  0.160 M NaCl = 0.160 M, The molarity in Cl⁻  0.180 M SrCl₂ = 0.360 M,  The molarity in Cl⁻ 0.150 M AlCl₃ = 0.450 M

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.

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