32 B. Molecular mass of an unknown acid. 1. Unknown number 2. Molarity of sodium

Question

32 B. Molecular mass of an unknown acid. 1. Unknown number 2. Molarity of sodium hydroxide solution Titrations flask 1 flask 2 flask 3 4 3. Mass of flask + unk. 132 84 132 328s 4. Mass of flask 5.533 5. Mass of unknown 6. Final buret reading Sm 8. Initial buret reading 9. Volume of NaOH soln.25ml nL 10. Moles of unknown acid (assume monoprotic) I1. Molecular mass of acid 12. Average molecular mass of acid 13. Deviation of each mol. mass from average Average deviation of molecular mass Suggest experimental modifications you could make if the acid to be titrated has 14. 15. solubility in water.

Explanation / Answer

Since the acid is a monoprotic acid, hence it will react in 1:1 with base NaOH

Flask 1:

Number of moles of NaOH = Volume of solution (in L) * Molarity (M)

=> 33.5/1000 * 0.01

=> 0.000335 moles

Moles of unknown acid = 0.000335 moles

Molecular mass of acid = Mass/number of moles = 5.58/0.000335 = 16656.71 gm/mol

Flask 2:

Number of moles of NaOH = Volume of solution (in L) * Molarity (M)

=> 43/1000 * 0.01

=> 0.00043 moles

Moles of unknown acid = 0.00043 moles

Molecular mass of acid = Mass/number of moles = 5.58/0.00043 = 12976.71 gm/mol

12) average molecular mass of acid = 1/2 * (16656.71 + 12976.71) = 14816.71

13) Deviation from the flask 1 = 1839.29

Deviation from the flask 2 = -1839.29

14) Average deviation = 1839.29

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