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Alex17521 [72]
2 years ago
10

What is the molarity of sodium ions in a solution prepared by diluting 250. mL of 0.550 M Na2SO4 to 1.25 L? (A) 0.110 M (C) 0.22

0 M (B) 0.138 M (D) 0.275 M
Chemistry
1 answer:
Leviafan [203]2 years ago
6 0

Answer:

(C) 0.220 M

Explanation:

Given parameters:

volume of the dillutant = 250mL = 0.25L

Molarity of Na₂SO₄ before being diluted = 0.55M

Volume of Na₂SO₄ after dilution = 1.25L

Unknown:

Molarity of Na⁺ ions in the diluted solution

Solution

We can simply adopt the mole concept in solving this dilution problem. During dilution, we intend to reduce the concentration of a particular solution by increasing its volume. Therefore, we have to find the new concentration of the diluted Na₂SO₄. From here, we can be able to find the molarity of the sodium ions in the  diluted solution:

      Using dilution equation:  \frac{C_{1} }{V_{1} } =  \frac{C_{2} }{V_{2} }

   we can solve for the molarity of the diluted solution

C₁ = concentration of the original solution

V₁ = volume of the original solution

V₂ = final volume of the diluted solution

C₂ = final volume

We are solving of C₂:

Making C₂ the subject of the formula gives:

                   C₂ = \frac{C_{1}V_{1} }{V_{2} }

           Therefore:

                    C₂ = \frac{0.25 x 0.55 }{1.25 } = 0.11M

The diluted concentration is now 0.11M

To find the molarity of Na⁺ in Na₂SO₄:

We write the ionic form of the compound in solution:

                            Na₂SO₄  → 2Na⁺ + SO₄²⁻

    In this solution,

           1 mole of Na₂SO₄ will produce 2 moles of Na⁺ ions

     From 0.11M of Na₂SO₄, we would have (2 x 0.11M) = 0.22M of Na⁺

The molarity of sodium ions will be 0.22M

               

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Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate
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The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: m_{(AcO)_2Cu} = 0.972 g

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Molarity of the sodium chromate solution: c_{Na_2CrO_4} = 0.0400 M

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

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Write the net ionic equation for this reaction:

Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

n_{(AcO)_2Cu} = 0.0053515 mol

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The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M

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