A 55.29-g sample of Ba(OH)2 is dissolved in enough water to make 1.20 L of solution. How many milliliters of this solution must be diluted with water in order to make 1.00 L of 0.100 M Ba(OH)2? (Ignore significant figures for this problem.)

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To solve this problem, we need to first calculate the initial concentration of the Ba(OH)2 solution, and then determine the volume needed to make the desired concentration.

Step 1: Calculate the initial concentration of the Ba(OH)2 solution.

We are given that we have a 55.29-g sample of Ba(OH)2 dissolved in enough water to make a 1.20 L solution. To find the concentration, we need to convert the mass of Ba(OH)2 to moles, and then divide by the volume of the solution.

The molar mass of Ba(OH)2 is calculated by adding up the atomic masses of its constituent elements:
1 Ba: 137.33 g/mol
2 O: 16.00 g/mol x 2 = 32.00 g/mol
2 H: 1.01 g/mol x 2 = 2.02 g/mol

Molar mass of Ba(OH)2 = 137.33 g/mol + 32.00 g/mol + 2.02 g/mol = 171.35 g/mol

Now, we can calculate the number of moles of Ba(OH)2:
moles of Ba(OH)2 = mass / molar mass = 55.29 g / 171.35 g/mol = 0.3223 mol

To find the initial concentration (Molarity), we divide the moles of solute by the volume of the solution:
initial concentration = moles / volume = 0.3223 mol / 1.20 L = 0.2686 M

Therefore, the initial concentration of the Ba(OH)2 solution is 0.2686 M.

Step 2: Determine the volume needed to make the desired concentration.

We are asked to find the volume of the Ba(OH)2 solution that needs to be diluted to make a 1.00 L solution with a concentration of 0.100 M.

The equation we can use is the dilution equation:
C1V1 = C2V2

C1 = initial concentration = 0.2686 M
V1 = volume of solution to be diluted (unknown)
C2 = desired concentration = 0.100 M
V2 = final volume = 1.00 L

To solve for V1, we rearrange the equation:
V1 = (C2V2) / C1 = (0.100 M) (1.00 L) / 0.2686 M ≈ 0.3726 L

To convert this volume to milliliters, we can multiply by 1000:
V1 = 0.3726 L * 1000 mL/L ≈ 372.6 mL

Therefore, approximately 372.6 mL of the Ba(OH)2 solution must be diluted with water in order to make 1.00 L of 0.100 M Ba(OH)2.