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What is the final concentration of 500 mL of a solution made from 125 mL of 3.0 M NaOH?

[tex]\[ \text{[?] M NaOH} \][/tex]


Sagot :

To find the final concentration of 500 mL of a solution made from 125 mL of 3.0 M NaOH, follow these steps:

1. Determine the moles of solute in the initial solution:
- The initial volume of the solution is 125 mL.
- The initial concentration is 3.0 M (moles per liter).
- First, convert the initial volume from mL to L: [tex]\( 125 \, \text{mL} = 0.125 \, \text{L} \)[/tex].

- Calculate the number of moles of NaOH in the initial solution using the formula:
[tex]\[ \text{moles of solute} = \text{concentration} \times \text{volume} \][/tex]
Substituting the values:
[tex]\[ \text{moles of solute} = 3.0 \, \text{M} \times 0.125 \, \text{L} = 0.375 \, \text{moles} \][/tex]

2. Calculate the final concentration after dilution:
- The final volume of the solution is 500 mL.
- Convert the final volume from mL to L: [tex]\( 500 \, \text{mL} = 0.500 \, \text{L} \)[/tex].

- Use the number of moles of solute (which remains unchanged during dilution) and the final volume to find the final concentration:
[tex]\[ \text{final concentration} = \frac{\text{moles of solute}}{\text{final volume}} \][/tex]
Substituting the values:
[tex]\[ \text{final concentration} = \frac{0.375 \, \text{moles}}{0.500 \, \text{L}} = 0.75 \, \text{M} \][/tex]

Therefore, the final concentration of the solution is:

[tex]\[ 0.75 \, \text{M NaOH} \][/tex]