6. A student used the table below to record the volume of NaOH dispensed whern n

Question

6. A student used the table below to record the volume of NaOH dispensed whern neutralizing 25.00 mL of HCl. Scout 1.10 23.50 Trial 1 0.70 22.80 22.10 Trial 2 22.80 44.85 22.05 Initial buret V (mL Final buret V (mL Total V dispensed (mL) 22.40 During the experiment the concentration of NaOH was given to the students and recorded in the table below. The volumes of NaOH come from the table above. The student used the data to calculate the concentration of HCI. Concentration of NaOH Volume of NaOH added (mL Moles of NaOH (mol) Volume of HCl in flask (mL Concentration of HC Trial 1 0.1225 MM 22.10 mL 0.002707 25.00 mlL 0.1083 M Trial 2 0.1225 MM 22.05 mL 0.002701 25.00 mlL 0.1081 MM After doing the calculations and reporting these values to their teaching assistant the students were told that the actual value for the concentration of HCl was 0.1083 M. Discuss the accuracy of the values that the student found for the concentration of HC.

Explanation / Answer

The actual value of concentration of HCl given = 0.1083M

The average value of concentration of HCl determined by the student = (trial1 value + trial 2 value)/ number of trials = (0.1083 + 0.1081) / 2 = 0.1082 M

Percentage error in calculation by the student = [(actual value- experimental value)/ actual value] x 100

[(0.1083- 0.1082)/0.1083] x 100 = 0.092%

Percentage accuracy = 100- percentage error = 100- 0.092 = 99.908%

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