What is the final concentration if water is added to 0.25 L of a 8M NaOH solution to make 4.0 L of a diluted NaOH solution? .25 M NaOH solution
1 M NaOH solution
.5 M NaOH solution

0.5 M NaOH solution

Explanation:

Initially, we have 0.25L or 250 mL of 8M NaOH solution

Which means that 1L of this solution would have 8 moles of NaOH

Moles present in 250 mL of solution:

Molarity * Volume( in L)

8 * 0.25 = 2 moles

There is 2 moles of NaOH in 0.25 L of the provided NaOH solution

More water is going to be added to this solution but the number of moles of NaOH will remain the same

now that the solution has a volume of 4L after adding the additional water, we have a 4L solution which contains 2 moles of NaOH

Molarity of the new solution:

Molarity = Number of moles in solution / Volume of solution

Molarity = 2 / 4

Molarity = 0.5 M

thats how its supposed to be

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