Calculate the volume of a 0.225M solution of KOH required to react with 0.215g of acetic acid.

V = 0.018 L

Explanation:

Given data:

Mass of acetic acid = 0.215 g

Molarity of KOH = 0.225 M

Volume of KOH required = ?

Solution:

Chemical equation:

CH₃COOH + KOH      →    CH₃COOK + H₂O

Number of moles of acetic acid:

Number of moles = mass/molar mass

Number of moles = 0.215 g / 60.0 g/mol

Number of moles = 0.004 mol

now we will compare the moles of acetic acid and KOH.

                          CH₃COOH       :       KOH

                                  1                :          1

                              0.004           :       0.004

Volume of KOH required:

Molarity = number of moles / volume

0.255 M = 0.004 mol / V

V = 0.004 mol / 0.225 mol/L

V = 0.018 L

10 years is the answer

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.