Respuesta :

Answer: The concentration of chloride ions in the solution obtained is 0.674 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]     .....(1)

  • For KCl:

Molarity of KCl solution = 0.675 M

Volume of solution = 297 mL

Putting values in equation 1, we get:

[tex]0.675=\frac{\text{Moles of KCl}\times 1000}{297}\\\\\text{Moles of KCl}=\frac{(0.675mol/L\times 297)}{1000}=0.200mol[/tex]

1 mole of KCl produces 1 mole of chloride ions and 1 mole of potassium ion

Moles of chloride ions in KCl = 0.200 moles

  • For magnesium chloride:

Molarity of magnesium chloride solution = 0.338 M

Volume of solution = 664 mL

Putting values in equation 1, we get:

[tex]0.338=\frac{\text{Moles of }MgCl_2\times 1000}{664}\\\\\text{Moles of }MgCl_2=\frac{(0.338mol/L\times 664)}{1000}=0.224mol[/tex]

1 mole of magnesium chloride produces 2 moles of chloride ions and 1 mole of magnesium ion

Moles of chloride ions in magnesium chloride = [tex](2\times 0.224)=0.448mol[/tex]

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.200 + 0.448) moles = 0.648 moles

Total volume of the solution = (297 + 664) mL = 961 mL

Putting values in equation 1, we get:

[tex]\text{Concentration of chloride ions}=\frac{0.648mol\times 1000}{961}\\\\\text{Concentration of chloride ions}=0.674M[/tex]

Hence, the concentration of chloride ions in the solution obtained is 0.674 M

mickymike92

Based on the concentrations of the given solutions, the concentration of Cl⁻ is 0.674 M.

What is the moles of chloride ion in each solution?

Moles of a substance is calculated using the formula:

  • Moles = molarity *  volume

For 297 mL of 0.675 M KCl

297 mL = 0.297 L

moles of KCl = 0.675 * 0.297

moles of KCl = 0.200 moles

1 mole of KCl produces 1 mole of Cl⁻

Thus, moles of Cl⁻ = 0.200 moles

For 664 mL of 0.338 M MgCl₂

664 mL = 0.664 L

moles of MgCl₂ = 0.338 * 0.664

moles of MgCl₂ = 0.224 moles

1 mole of MgCl₂ produces 2 moles of Cl⁻

moles of Cl⁻ = 2 * 0.224

moles of Cl⁻ = 0.448 moles

Total moles of Cl⁻ = 0.448 + 0.200

moles of Cl⁻ = 0.648 moles

Volume of solution = 664 + 297

Volume of solution = 961 = 0.961 L

Molarity of Cl⁻ = 0.648 / 0.961

Molarity of Cl⁻ = 0.674 M

Therefore, the concentration of Cl⁻ is 0.674 M

Learn more about molarity at: https://brainly.com/question/489225

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