This is a dilution problem involving molarity. The formula to use is:
#M_1V_1=M_2V_2#,
where #M# is molarity and #V# is volume in liters.
The unit for molarity is #"mol solute/L of solution"#, or simply #"mol/L"#. So a 2.5 M solution contains #"2.5 mol solute/L of solution"#. Therefore, all volumes must be in liters. The final volume can be converted into milliliters, as is the case in this question.
Organize the data:
Known
#M_1="6.0 M HCl"##=##"6.0 mol/L HCl"#
#M_2="0.50 M HCl"##=##"0.50 mol/L HCl"#
#V_2=100.0 color(red)cancel(color(black)("mL"))xx(1"L")/(1000color(red)cancel(color(black)("mL")))="0.1000 L"#
Unknown
#V_1#
Solution
Rearrange the formula to isolate #V_1#. Insert the data into the equation and solve.
#V_1=(M_2V_2)/M_1#
#V_1=(0.50color(red)cancel(color(black)("M"))xx0.1000"L")/(6.0color(red)cancel(color(black)("M")))="0.0083 L"#
(rounded to two significant figures)
Convert liters to milliliters.
#0.0083color(red)cancel(color(black)("L"))xx(1000"mL")/(1color(red)cancel(color(black)("L")))="8.3 mL"#
