Respuesta :
Answer:
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.
