The first step that you will do in lab is to prepare approximately 20.0 mL each of the following solutions of NaOH, using the NaCl solution to dilute the concentrated (0.300 M) solution of NaOH. After you prepare each solution, transfer it to a clean 50 mL beaker on top of a labeled piece of paper. 0.200 M NaOH 0.150 M NaOH 0.100 M NaOH 0.050 M NaOH 0.025 M NaOH

0.200 M NaOH: 13.3 mL of the concentrated solution and complete to 20.0mL

0.150 M NaOH: 10.0 mL of the concentrated solution and complete to 20.0mL

0.100 M NaOH: 6.67 mL of the concentrated solution and complete to 20.0mL

0.050 M NaOH: 3.33 mL of the concentrated solution and complete to 20.0mL

0.025 M NaOH: 1.67 mL of the concentrated solution and complete to 20.0mL

Explanation:

It is possible to prepare a solution from a more concentrated one. In the problem, the concentrated solution is 0.300M NaOH. Thus, to prepare 20.0mL of each of the solutions you will need:

0.200 M NaOH: 20.0mL × (0.200M / 0.300M) = 13.3 mL of the concentrated solution and complete to 20.0mL

The ratio between the concentrated solution and the solution you want to prepare is called "dilution factor"

0.150 M NaOH: 20.0mL × (0.150M / 0.300M) = 10.0 mL of the concentrated solution and complete to 20.0mL

0.100 M NaOH: 20.0mL × (0.100M / 0.300M) = 6.67 mL of the concentrated solution and complete to 20.0mL

0.050 M NaOH: 20.0mL × (0.050M / 0.300M) = 3.33 mL of the concentrated solution and complete to 20.0mL

0.025 M NaOH: 20.0mL × (0.025M / 0.300M) = 1.67 mL of the concentrated solution and complete to 20.0mL

i believe it is 40–40–20%

explanation: