Will mark BRAINLIEST. Molarity
Please no Bs answers. Only going to be reported.

If water is added to 145 mL of a 0.55 M KOH solution until the volume is 250 mL, what will the molarity of the diluted solution be?

What is the molarity of the solution that results from diluting 35.0 ml of a 9.02M solution to a new volume of 45.0 ml?

where M is the molarity, V is the volume, and 1 and 2 refer to the initial and final states of the solution, respectively. Here, we are given the molarity of the initial solution and the volumes of the initial and final (diluted) solutions. To find the final concentration (i.e., the molarity of the diluted solution), we would be solving for M₂:

[tex]M_2=\frac{M_1V_1}{V_2} = \frac{(0.55 \text{ M})(145 \text{ mL})}{250 \text{ mL}} \\ M_2 = 0.319 \approx 0.32 \text{ M}.[/tex]

The molarity is given to two significant figures as both our M₁ and V₂ are given to two significant figures.

Note: Although our volumes are in mL instead of L, we do not need to convert them to L for the purposes of our calculation since we would be multiplying our V₁ and V₂ by a common factor that would cancel out in division. All that matters is the ratio between the two volumes, which is the same whether the volumes are in mL or in L.

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We follow the exact same procedure in the second question as we did in the first problem: solve for M₂ given M₁ = 9.02 M, V₁ = 35.0 mL, and V₂ = 45.0 mL:

[tex]M_2=\frac{M_1V_1}{V_2} = \frac{(9.02 \text{ M})(35.0 \text{ mL})}{45.0 \text{ mL}} \\ M_2 = 7.02 \text{ M}.[/tex]