Calculato the volume in milliters of a 0.121M barium acetate solution that contains 25.0 g of barlum acetate (Ba(C2H3O2)2)Be sure your answer has the correct number of significant digits.

Answer 1

To find the volume of the solution, we first need to calculate the number of moles of barium acetate present in 25.0 g.

Molar mass of Ba(C2H3O2)2 = 137.33 g/mol + 2(12.01 g/mol + 1.008 g/mol + 16.00 g/mol) = 255.37 g/mol

Number of moles = mass / molar mass = 25.0 g / 255.37 g/mol = 0.098mols

Now, we use the definition of molarity to find the volume of the solution:

Molarity (M) = moles of solute / volume of solution (L)

0.121 M = 0.098 moles / volume (L)

Volume (L) = 0.098 moles / 0.121 mol/L = 0.809 L

Lastly, we convert this volume to milliliters:

1 L = 1000 mL

0.809 L x 1000 mL/L = 809 mL

Therefore, the volume of the 0.121 M barium acetate solution that contains 25.0 g of barium acetate is 809 mL.