M1V1=M2V2 The simple dilution equation. You can use this anytime you want to dilute another solution (that is, make a solution from another solution of the same thing just at a different concentration). so we have the molar mass of H2C2O4 as being 1.008(2)+12.01(2)+15.999(4)=2.016+24.02+63.996=90.032 grams per mol of H2C2O4 That means that if I were to put 90.032 grams of this compound in 1 L of water I would have a 1M solution. instead I am putting 72.4g so I take 72.4g(1mol/90.032g) = 0.804 mol of H2C2O4 if I put that in 1 L of H2O I have (0.804mol/1L)= a 0.804M solution of H2C2O4 notice the units now if I have a 0.804M sol and I need to make a 0.145M one I use the dilution equation. M1 is the first concentration so M1=0.804 M2 is the desired concentration so M2=0.145 V1 is the amount of 0.804M sol I would add SO this is what I will be solving for V2 is the amount of 0.145M solution I want to make. so take (0.804M)(x L)=0.145M(1L) so x Liters I will need. That makes (0.145M(1L))/(0.804M)=0.180L of the 0.804M solution to make 1L of a 0.145M sol. Converted to mL that would be 180mL. so by filling in the blanks you gave me: Place _180__ mL of the __0.804__ mole solution in a 1 L volumetric flask, and dilute to 1.00 L. Hope that helps I tried to be very through.