I tried by my self but I don’t sure if the are correct! Also need help we the la

Question

I tried by my self but I don’t sure if the are correct!
Also need help we the last two Molarity of NaoH titrant solutionQ 6093 Molar Mass of Acetic Acid, CHCOOH GO.03m a) Mass of empty 125 mlL Volume of vinegar used for each trial 0.Q0 mL which is 0l00 Trial1Trial 2Trial 3ria nhs333 131 36 0g lask (g) b) Mass of 125 mL flask with vinegar (g) o) Mass of vinegar () 10.02 .93 9.98s d) NaOH- initial burette reading (mL) ynl. reading (mL) f) Volume of NaOH used to titrate sample (mL) g) Volume of NaOH used 6.01922 0109L 00880.0095 000835 L) h) Moles of NaOH used to itrate sample i) Moles of acetic acid in sample j) Molarity of acetic acid in vincgar k) Mass of acetic acid in sample (g I) % (m/m) acetic acid in vinegar sample .536g 0.522 0-55g % Deviatio Standard Deviation Mean Molarity of acetic acid in 29 001G4 .2433 6.0334 m/m of acetic acid in vinegar

Explanation / Answer

Calculate molarity of acetic acid in vinegar solution

All the calculations upto l (first table) are correct

Second table

Standard deviation (molarity acetic acid) = 0.0002 M

Standard deviation (%m/m) in vinegar = 0.051

So,

third column

x+/-s becomes molarity acetic acid = 0.8729 +/- 0.0002 M

x+/-s becomes %m/m acetic acid in vinegar = 5.2433 +/- 0.051

% deviation,

molarity acetic acid = 0.0002 x 100/0.8729 = 0.023%

%m/m acetic acid in vinegar = 0.051 x 100/5.2433 = 0.973%

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