What is the final concentration if water is added to 0.50 l of a 12 m naoh solution to make 3.0 l of a diluted naoh solution?

1 m naoh solution

2 m naoh solution

3 m naoh solution

4 m naoh solution

The answer to your question is the second option: 2M NaOH solution

Explanation:

Data

Volume 1 = 0.5 l

Concentration 1 = 12 M NaOH

Volume 2 = 3 l

Concentration 2 = ?

Formula

                     V₁C₁  =  V₂C₂

Solve for C₂

                     C₂ = V₁C₁ / V₂

Substitution

                    C₂ = (0.5)(12) / 3

Simplification and result

                    C₂ = 6 / 3

                    C₂ = 2 M  

2 M NaOH solution

Explanation:

Initial concentration of NaOH = 12 M

Initial volume of solution = 0.5

Initial number of moles = Molarity×Volume = 12×0.5 = 6 moles

Note: On dilution the number of moles does not change.

Initial number of moles = final number of moles.

Final number of moles = 6 moles

Final volume of solution = 3 L

Final concentration (in Molarity) =

radioactive decay occurs in unstable atomic nuclei – that is, ones that don't have enough binding energy to hold

the nucleus together due to an excess of either protons or neutrons.

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.