Answer: A volume of 600 mL of 3.0 M [tex]H_{2}SO_{4}[/tex] solution can be prepared by using 100.0 mL OF 18 M [tex]H_{2}SO_{4}[/tex].
Explanation:
Given: [tex]V_{1}[/tex] = ?, [tex]M_{1} = 3.0 M\\[/tex]
[tex]V_{2} = 100.0 mL[/tex], [tex]M_{2} = 18 M[/tex]
Formula used to calculate the volume 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}\\3.0 M \times V_{1} = 18 M \times 100.0 mL\\V_{1} = \frac{18 M \times 100.0 mL}{3.0M}\\= 600 mL[/tex]
Thus, we can conclude that a volume of 600 mL of 3.0 M [tex]H_{2}SO_{4}[/tex] solution can be prepared by using 100.0 mL OF 18 M [tex]H_{2}SO_{4}[/tex].