Question

A chemist adds 240.0 mL of a 0.97 mol/L barium acetate (BaC,H,02), ) solution to a reaction flask. Calculate the millimoles of barium acetate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

Answer 1

we are given the volume of Barium acetate solution = 240.omL -0.24L (Десаме 1L - ооо чL - so ImL = /1000 L - This 240.0mL = 240.0 = 0.24.lt 1000 and molar concentration is 0.97 mol/L. formula of u b of molarity عهد We will use the to calculate moles in the solution. The formula is acetate Molarity = Number of moles of solute Volume of solution in a ملے۔ No. of moles = (Melarity) xf volume of solutia . in ) Here solute is Barium Acetate Thus (Molarity or Molar No. of moles - -concenthation of Barius . acetate) x (volume a of Blarium acetate soluti - (0.97 mol/&% (0.24x) = 0.92x0.24 mol=0:23

- -- Now, As I milli (m) = 10-3 units sa millindle = 10-3 mole or millimale = imole 10-3 Thus I mole = = 103 millimole - 10-3 Thus we can convert moles to millimoles by multiplying with 105.

so ace Number of moles of barinn 0.23 moede of Bassim - - 10.23 X 10² millimoles Thus added the millimoles of Barium acetate to the flask are 0 23 X 10² mmol