Starting with a 6.00m stock solution of sodium sulfate four so called standard solutions (1-4) are prepared by sequentially diluting 10.00 ml of each solution to 250.00 ml. for instance solution 1 is prepared by diluting the stock solution, solution 2 is prepared by diluting solution 1, etc. determine the concentration of all standard solutions and the number of moles of na⁺ ions in each solution.

Explanation:

This is dilution problem.

As given that we have stock solution of concentration 6 M.

We have taken 10mL of this stock solution to form a new solution with volume 250 mL.

We can use

MV = M₁V₁

Where M = concentration of stock solution = 6 M

V = volume of stock solution = 10 mL

M₁ = concentration of dilute solution (after first dilution) = ?

V₁ = final volume = 250 mL

Putting values

6 X 10 = M₁X250

M₁ = 0.24 M

The formula of sodium sulfate is Na₂SO₄

So each moles of sodium sulfate will have two moles of sodium ion.

Hence the moles of sodium ion in standard solution 1 = 2 X 0.24 mol /L =0.48 moles / L.

2)  M₁V₁ = M₂V₂

Where M₁ = concentration of standard solution 1 = 0.24 M

V₁ = volume of standard solution 1 = 10 mL

M₂ = concentration of dilute solution (after second dilution) = ?

V₂ = final volume = 250 mL

Putting values

0.24 X 10 = M₂X250

M₂= 0.0096 M

The formula of sodium sulfate is Na₂SO₄

So each moles of sodium sulfate will have two moles of sodium ion.

Hence the moles of sodium ion in standard solution 2 = 2 X 0.0096 mol /L = 0.0192 moles / L.

3)

M₂V₂ = M₃V₃

Where M₂ = concentration of standard solution 2 = 0.0096 M

V₂ = volume of standard solution 2 = 10 mL

M₃ = concentration of dilute solution (after third dilution) = ?

V₃ = final volume = 250 mL

Putting values

0.0096 X 10 = M₃ X 250

M₃ = 0.000384 M

The formula of sodium sulfate is Na₂SO₄

So each moles of sodium sulfate will have two moles of sodium ion.

Hence the moles of sodium ion in standard solution 3 = 2 X 0.000384 mol /L = 0.000768 moles / L.

4)

 M₃V₃ = M₄V₄

Where M₃ = concentration of standard solution 3 = 0.000384 M

V₃ = volume of standard solution 3 = 10 mL

M₄ = concentration of dilute solution (after fourth dilution) = ?

V₄ = final volume = 250 mL

Putting values

0.000384 X 10 = M₄ X 250

M₄ = 0.00001536 M

The formula of sodium sulfate is Na₂SO₄

So each moles of sodium sulfate will have two moles of sodium ion.

Hence the moles of sodium ion in standard solution 4 = 2 X 0.00001536 mol /L = 0.00003072 moles / L.

b is the correct