What is the final molarity of KOH in the 100. mL flask ?

Question

A student prepares a solution by dissolving 3.75 g of KOH in enough water to make 250. mL of solution.
The student then transfers 27.0 mL of this solution into a 100. mL volumetric flask. Water is then added
up to the 100. mL mark.

Answer 1

The number of moles of KOH used is:

3.75 g / 56.1 g/mol = 0.0668 moles

in 250 mL (1/4 L) this is 0.0688 * 4 = 0.267 M

M1 * V1 = M2 * V2 or M2 = (M1 * V1)/V2

= (0.267 M * 27mL)/100mL = 7.209 / 100

= 0.0721 M