Include copies of all tables generated in this experiment. Add a line to each of
ID: 510846 • Letter: I
Question
Include copies of all tables generated in this experiment. Add a line to each of these tables that gives the volume of the reagent dispensed during the titration. Include copies of all calculations based on the data collected in these tables. Use the that one drop is 0.05 mL to estimate the accuracy with which the concentration of the NaOH solution can be determine in Part I. Compare the results of Part III in which the unknown acid solution was standardized with NaOH with the results obtained when this solution was standardized with KMno_4. Assume that you wanted to prepare 250 mL of a 0.100 M H_2C_22O_4 solution. Oxalic acid is available as a white crystal with the formula H_2C_2O_4 C 2 H_20. Calculate the mass of the oxalic acid needed to prepare this solution. Explain why it isn't possible to prepare a 0.100 M H_2C_2O_4 solution by adding 0.100 moles of oxalic acid to exactly one liter of water.Explanation / Answer
Answer:
1) Volume of Oxalic acid = 250 mL
Molarity of Oxalic acid = 0.1 M
H2C2O4. 2H2O molar mass = 126
Molarity = (weight / molar mass) x (1000 / V in mL)
0.1 = (wt / 126) x (1000 / 250)
0.1 = (wt / 126) x 4
0.1 x 126 = 4 x wt
4 x wt = 12.6
wt = 12.6 / 4
wt = 3.15 g.
Hence the mass of Oxalic acid to prepare the solution is 3.15 g
2) Here the concentration is caluculating in Molarity.
According to molarity principle, The number of moles of solute dissolved in one liter of the solution.
in the question 0.1 moles given for the one liter, So its Molarity becomes 0.1 M