Question

B. Determination of the Molarity of Unknown Acid Trial 1 Trial 2 Trial 3 Mass of Erlenmeyer flask (g) 124 g Mass of Erlenmeyer flask+acid After adding 5 mL HCl solution (g) 135 g Mass of HCl solution (g) Mass of HCl solution (kg) -------- ------ -- Volume (acid), mL ----5mL------- -----5 mL--- ------5mL---- NaOH Initial Volume (Vi) 0.00 mL ------------mL -------------mL NaOH final Volume (VA) 16.5 mL ----------- -------------m L ml ---------mL NaOH Volume Used Vi-Vi --------mL L

Answer 1

The mass of HCl is calculated:

m HCl = 135-124 = 11 g

In Kg:

m HCl = 11 g * (1 Kg / 1000 g) = 0.011 Kg

The volume of NaOH is calculated:

V NaOH = 16.5 - 0 = 16.5 mL

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