A0.20 mol sample of mgcl2(s) and a 0.10 mol sample of kcl(s) are dissolved in water and diluted to 500 ml. what is the concentration of cl- in the solution?

1 M

Explanation:

Magnesium chloride will furnish chloride ions as:

Given :

Moles of magnesium chloride = 0.20 mol

Thus, moles of chlorine furnished by magnesium chloride is twice the moles of magnesium chloride as shown below:

Moles of chloride ions by magnesium chloride = 0.40 moles

Potassium chloride will furnish chloride ions as:

Given :

Moles of potassium chloride = 0.10 moles

Thus, moles of chlorine furnished by potassium chloride is same as the moles of potassium chloride as shown below:

Moles of chloride ions by potassium chloride = 0.10 moles

Total moles = 0.40 + 0.10 moles = 0.50 moles

Given, Volume = 500 mL = 0.5 L (1 mL = 10⁻³ L)

Concentration of chloride ions is:

The final concentration of chloride anion = 1 M

The concentration of the chloride ion, Cl¯ in the solution is 1 M

To obtain the answer to the question given above, we'll begin by calculating the number of mole of chloride ion, Cl¯ produced by each compound in the solution. This can be obtained as follow:

MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)

From the balanced equation above,

1 mole of MgCl₂ produced 2 moles of Cl¯.

Therefore, 0.2 mole of MgCl₂ will produce = 0.2 × 2 = 0.4 mole of Cl¯

KCl(aq) —> K⁺(aq) + Cl¯(aq)

From the balanced equation above,

1 mole of KCl produced 1 mole of Cl¯

Therefore,

0.1 mole of KCl will also produce 0.1 mole of Cl¯.

Next, we shall determine the total mole of Cl¯ in the solution.

Mole of Cl¯ from MgCl₂ = 0.4 mole

Mole of Cl¯ from KCl = 0.1 mole

Total mole of Cl¯ = 0.4 + 0.1

Finally, we shall determine the concentration of Cl¯ in the solution.

Total mole of Cl¯ = 0.5 mole

Volume = 500 mL = 500 / 1000 = 0.5 L

Concentration = mole / Volume

Concentration of Cl¯ = 0.5 / 0.5

Therefore, the concentration of Cl¯ in the solution is 1 M

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explanation:

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