Answer:
[tex]V_1=2.49mL[/tex]
Explanation:
Hello,
In this case, considering that the moles of hydrioiodic acid remain unchanged during the dilution process:
[tex]n_{HI}=n_{HI}[/tex]
One could apply the following equality in terms of molarity:
[tex]V_1M_1=V_2M_2[/tex]
Whereas the subscript 1 accounts for the solution before the dilution and 2 after the dilution, therefore, the required volume of 6.00 M acid is:
[tex]V_1=\frac{V_2M_2}{M_1} =\frac{50.0mL*0.299M}{6.00M}=2.49mL[/tex]
Best regards.