A 51.24 g sample of Ba(OH)2 is dissolved in enough water to make 1.20 L of solution. How many mL of this solution must be diluted with water to make 1.00 L, of 0.100 M Ba(OH)2?

You must dilute 401 mL with enough water to make 1.00 L of solution.
Step 1. Calculate the of the stock solution.
Moles of Ba(OH)₂ = ##”51.24 g Ba(OH)”_2 × (“1 mol Ba(OH)”_2)/(“171.34 g Ba(OH)”_2) = “0.299 05 mol Ba(OH)”_2##
Molarity = ##”moles”/”litres” = “0.299 05 mol”/”1.20 L”## = 0.2492 mol/L
Step 2. Calculate the volume of stock solution that is needed.
See
##c_1V_1 = c_2V_2##
##c_1## = 0.2492 mol/L; ##V_1## = ?
##c_2## = 0.100 mol/L; ##V_2## = 1.00 L
##V_1 = V_2 × c_2/c_1## = ##”1.00 L” × “0.100 mol/L”/”0.2492 mol/L”## = 0.401 L = 401 mL
So, to make your solution, you would dilute 401 mL of stock solution with enough water to make 1.00 L.