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) 0.220 m (b) 0.138 m (d) 0.275 m

(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:   =  

   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 =

           Therefore:

                    C₂ = = 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

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.

the answer to your question is c