Answer: A volume of 500 mL water is required to prepare 0.1 M [tex]H_{3}PO_{4}[/tex] from 100 ml of 0.5 M solution.
Explanation:
Given: [tex]M_{1}[/tex] = 0.1 M, [tex]V_{1}[/tex] = ?
[tex]M_{2}[/tex] = 0.5 M, [tex]V_{2}[/tex] = 100 mL
Formula used to calculate the volume of water is as follows.
[tex]M_{1}V_{1} = M_{2}V_{2}[/tex]
Substitute the values into above formula as follows.
[tex]M_{1}V_{1} = M_{2}V_{2}\\0.1 M \times V_{1} = 0.5 M \times 100 mL\\V_{1} = 500 mL[/tex]
Thus, we can conclude that a volume of 500 mL water is required to prepare 0.1 M [tex]H_{3}PO_{4}[/tex] from 100 ml of 0.5 M solution.
