Use the molarity dilution equation.
#M_1V_1=M_2V_2,#
where:
#M_1# is the molarity of the stock solution, #V_1# is the volume of the stock solution, #M_2# is the molarity of the diluted solution, and #V_2# is the volume of the diluted solution.
Known
#M_1="2.00 M"="2.00 mol/L"#
#V_1=25color(red)cancel(color(black)("mL"))xx(1"L")/(1000color(red)cancel(color(black)("mL")))="0.025 L"#
#M_2="0.200 M"="0.200 mol/L"#
Unknown
#V_2#
Solution
Rearrange the equation to isolate #V_2#. Plug in the known values and solve.
#V_2=(M_1V_1)/(M_2)#
#V_2=(2.00color(red)cancel(color(black)("mol"/"L"))xx0.025"L")/(0.200color(red)cancel(color(black)("mol"/"L")))="0.25 L"#
#V_2=0.25color(red)cancel(color(black)("L"))xx(1000"mL")/(1color(red)cancel(color(black)("L")))="250 mL"#
The final volume of the diluted #"MgSO"_4"# solution is #"0.25 L = 250 mL"#.
