Answer:Molarity of (MOM ) = 0.048 M
Explanation :
In this reaction , HCL acts as an acid and MOM act as a base
Given :
• Acid Concentration HCL = 0.2 M
,
• Average volume of HCl = 24 mL
,
• Volume Base (MOM) = 100 mL
,
• Concentration (MOM) = ?M
We will calculate the concentration of MOM by applying the dilutions formula :
[tex]M_{acid\text{ }}*\text{ V}_{acid\text{ }}=\text{ M}_{base\text{ }}*\text{ V}_{base\text{ }}[/tex]
Substituting the given parameters into the formula above, we calculate that Molarity (concentration) of MOM is :
[tex]\begin{gathered} M_{base\text{ }}=\frac{M_{acid}*\text{ V}_{acid\text{ }}}{V_{base\text{ }}} \\ \text{ = }\frac{0.2\text{ M * 24 mL }}{100\text{ mL }} \\ \text{ =0.048 M } \end{gathered}[/tex]
This means that Molarity of (MOM ) = 0.048 M
Post lab question 1 :
Balanced reaction as follows :
[tex]2HCl\text{ + Mg\lparen OH\rparen}_2\Rightarrow MgCl_2+2H2O\text{ }[/tex]
