A solution is prepared by dissolving 0.6901 g oxalic acid (H2C2O4) in enough water to make 100.0 mL of solution. A 10.00 mL aliquot (portion) of this solution is then diluted to a final volume of 250.0 mL. What is the final molarity of the oxalic acid solution?
And PLEASE explain how. Thank you!
General Chemistry help---Dilution Solution Problem?
2016-06-13 9:40 am
回答 (1)
2016-06-13 9:57 am
Molar mass of H₂C₂O₄ = (1.008×2 + 12.01×2 + 16.00×4) g/mol = 90.04 g/mol
No. of moles of H₂C₂O₄ = (0.6901 g) / (90.04 g/mol) = 0.007664 mol
Molarity of H₂C₂O₄ = (0.007664 mol) / (100.0 mL) = 0.07664 M
Dilution :
M₁V₁ = M₂V₂
(0.07664 M) × (10 mL) = M₂ × (250 mL)
Final molarity of H₂C₂O₄ = (0.07664 M) × (10/250) = 0.003066 M
No. of moles of H₂C₂O₄ = (0.6901 g) / (90.04 g/mol) = 0.007664 mol
Molarity of H₂C₂O₄ = (0.007664 mol) / (100.0 mL) = 0.07664 M
Dilution :
M₁V₁ = M₂V₂
(0.07664 M) × (10 mL) = M₂ × (250 mL)
Final molarity of H₂C₂O₄ = (0.07664 M) × (10/250) = 0.003066 M
收錄日期: 2021-04-18 15:00:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160613014056AA6tZwS