Lena772
  • Lena772
Carboxylic acids are organic acids that contain the -COOH group. A monoprotic Carboxylic Acid (0.5899 g) is neutralized with 48.4 mL 0.0998 M NaOH. What is the Molar Mass (in grams / mole) of the Carboxlyic Acid?
Chemistry
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Lena772
  • Lena772
48.4mL = 0.0484 L 0.484 L * 0.0998 M = mol solute 0.0483032 = mol solute MM= g/mol 0.5899g/ 0.0483032mol = 12.21 g/mol. I got it wrong however.
Lena772
  • Lena772
@Photon336 @Elsa213 @aaronq
Elsa213
  • Elsa213
"Ste[p 1: a balanced equation. HCOOH + NaOH -----------------> NaCOO + HOH Step 2: Write what you know under the equation. XCOOH + NaOH -----------------> NaCOO + HOH m = 0.5899 g v = 48.4 mL M = ? C = 0.0998 mol/L Step 3: Convert the Vol and Conc into moles of NaOH moles of NaOH = 0.0998 mol/L x 48.4 mL = 4.83 millimoles STep 4: Compare the moles of -COOH to moles of NaOH from the equation. It is 1 to 1 so every mole of NaOH needs the same number of moles of -COOH therefore moles of -COOH is 4.83 millimoles. Step 5: Now that you have a mass and number of moles of -COOH you can calculate M M = grams / moles 0.5899 g / 0.00483 mol = 122 .13 g/mol"

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Elsa213
  • Elsa213
https://answers.yahoo.com/question/index?qid=20111120183255AAWLaFY ;o
Elsa213
  • Elsa213
Hopefully that helped :3

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