Question

Sample 1 .674a Sample 3 0.00 -63619_573989 Sample 2 Mass of KHP taken Initial NaOH buret reading c om m e Final NaOH buret reading 33. ML 32.30ml Volume of NaOH used 33 mL 32.3mL Moles of KHP present 003_ moles_003 mole s kup Molarity of NaOH solution 10 M Mean molarity of NaOH solution and average deviation 0 04 M اور * 5% vinesar Sample 2 WHO Analysis of Vinegar Solution (Part 4a) Identification number (or brand) of vinegar sample used Sample 1 Quantity of vinegar taken 5.00mL Initial NaOH buret reading 0.00ml Final NaOH buret reading 40.60 ml Volume of NaOH used 40.6 mL Molarity of vinegar Mean molarity of vinegar and average deviation 7 % by mass acetic acid present ?

Answer 1

The average molar concentration of NaOH is 0.1 M.

Vinegar molarity is calculated:

M = Cb * Vb / Va = 0.1 M * 40.6 mL / 5 mL = 3.24 M

Since there was only one vinegar titration, the average is equal to 3.24 M

The percentage by mass is calculated:

% m / m = 3.24 mol / L * (60.05 g / 1 mol) * (1 L / 1000 mL) * 100 = 19.5%

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