Answer: There are 0.0637 moles present in 85.0 mL of 0.750 M KOH.
Explanation:
Given: Volume = 85.0 mL (1 mL = 0.001 L) = 0.085 L
Molarity = 0.750 M
It is known that molarity is the number of moles of solute present in liter of a solution.
Therefore, moles present in given solution are calculated as follows.
[tex]Molarity = \frac{moles}{Volume (in L)}\\0.750 M = \frac{moles}{0.085 L}\\moles = 0.0637 mol[/tex]
Thus, we can conclude that there are 0.0637 moles present in 85.0 mL of 0.750 M KOH.
