A chemist adds 125.0mL of a 0.77M barium acetate BaC2H3O22 solution to a reaction flask. Calculate the mass in grams of barium acetate the chemist has added to the flask. Round your answer to 2 significant digits.
Question
A chemist adds 125.0mL of a 0.77M barium acetate BaC2H3O22 solution to a reaction flask. Calculate the mass in grams of barium acetate the chemist has added to the flask. Round your answer to 2 significant digits.
Solution
To calculate the mass of barium acetate added to the flask, we need to use the formula:
Molarity (M) = moles of solute / volume of solution in liters
First, we need to convert the volume of the solution from mL to L:
125.0 mL = 0.125 L
Next, we rearrange the formula to solve for moles of solute:
moles of solute = Molarity (M) * volume of solution in liters
moles of barium acetate = 0.77 M * 0.125 L = 0.09625 moles
Finally, we convert moles of barium acetate to grams using the molar mass of barium acetate (Ba(C2H3O2)2), which is approximately 255.42 g/mol:
mass of barium acetate = moles of barium acetate * molar mass of barium acetate
mass of barium acetate = 0.09625 moles * 255.42 g/mol = 24.59 g
So, the chemist has added approximately 24.59 grams of barium acetate to the flask. However, we need to round our answer to 2 significant digits, so the final answer is 25 g.
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