Question

Mn2+

Answer 1

The molarity of the stock Mn²⁺ ions is 0.0288 M

Based on the dilution formula;

• The molarity of A is 0.00144 M
• The molarity of B is 0.0000576 M
• The molarity of C is 0.000001152 M

What is the molarity of a solution?

The molarity of a solution is the number of moles of a solute dissolved in a given volume of solution in liters.

• Molarity = number of moles/volume

The molarity of the stock solution is:

moles of Mn²⁺ ions = mass / molar mass

molar mass of  Mn²⁺ ions = 55.0 g/mol

moles of Mn²⁺ ions = 1.584 / 55

moles of Mn²⁺ ions = 0.0288 moles

molarity of Mn²⁺ ions = 0.0288 / 1

molarity of Mn²⁺ ions = 0.0288 M

The dilution formula is used to determine the molarities of A, B, and C.

C₁V₁ = C₂V₂

C₂ = C₁V₁ / V₂

Where;

• C₁ = initial molarity
• V₁ = initial volume
• C₂ = final molarity
• V₂ = final volume

Molarity of A = 50 * 0.0288 / 1000

Molarity of A = 0.00144 M

Molarity of B = 10 * 0.00144 / 250

Molarity of B = 0.0000576 M

Molarity of C = 10 * 0.0000576 / 500

Molarity of C = 0.000001152 M

Learn more about molarity at: brainly.com/question/17138838

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