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Unformatted text preview: Trial 1 Trial 2 Total volume NaOH 8.9mL 8.6mL Molarity, M, of standardized HCL 0.1M Molarity, M, in standardized HCL 0.1M By Equation 3, .1*25/8.9 .1*25/8.6 Trial1 Trial 2 M OH .2809M .2907M Average molarity of NaOH solution, M .2858M B. Determination of the Molar Mass of the Unknown Acid Trial 1 Trial 2 Mass of sample .504g .503g Volume NaOH used 9.4mL 9.5mL (1)No. moles NaOH=(V*M)/1000 (9.4*.2858)/1000 .0027M .0027M (9.5*.2858)/1000 No. moles H+ in sample---same as (1) .0027M .0027M MM=(no. grams acid)/(no. moles H+) .504/.0027 186.67g .503/.0026186.3g Unknown no. _B_...
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- Fall '07