Answer: 4.8 mL of 1.043 M acetic acid that is needed to prepare 50.0 mL of 0.10 M acetic acid
Explanation:
According to the dilution law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock [tex]CH_3COOH[/tex] solution = 1.043 M
[tex]V_1[/tex] = volume of stock [tex]CH_3COOH[/tex] solution = ?
[tex]M_1[/tex] = molarity of diluted [tex]CH_3COOH[/tex] solution = 0.10 M
[tex]V_1[/tex] = volume of diluted [tex]CH_3COOH[/tex]solution = 50.0 ml
Putting in the values we get:
[tex]1.043\times V_1=0.10\times 50.0[/tex]
[tex]V_1=4.8ml[/tex]
Therefore, volume in mL of 1.043 M acetic acid (CH3COOH) that is needed to prepare 50.0 mL of 0.10 M acetic acid is 4.8
