Question

12. Describe how to prepare a 250mL 0.350M of oxalic acid solution using oxalic acid dihydrate solids (H2C204.2H20, molar mass = 126.06g/mol) You need to describe appropriate glassware in your preparation. 13. Describe how you would prepare 500. mL of 0.10M of HCl solution from 3.0M of HCI stock solution. You need to describe appropriate glassware in your preparation.

Answer 1

12 ) Here we need to calculate mass of oxalic acid dihydrate required to prepare 250 ml of 0.350 M oxalic acid solution. To calculate mass , we will first calculate no. of moles & from no. of moles we will get mass of oxalic acid dihydrate required.

We know that, [ H₂C₂O₄.2H₂O ]= No. of moles of H₂C₂O₄.₂H₂O / volume of solution in L

$\therefore$ No. of moles of H₂C₂O₄.2H₂O = [ H₂C₂O₄.₂H₂O ] $\times$ volume of solution in L

$\therefore$ No. of moles of H₂C₂O₄.2H₂O = 0.350 mol / L $\times$ 0.250 L = 0.0875 mol

We know that, no. of moles = Mass / Molar mass

$\therefore$ Mass = No. of moles $\times$ molar mass

$\therefore$ Mass of H₂C₂O₄.2H₂O = 0.0875 mol $\times$ 126.06 g / mol = 11.0 g

Procedure : Weigh out accurately 11.0 g H₂C₂O₄.2H₂O. Transfer this solid into 250 ml beaker. Dissolve it in minimum amount of distilled water. Transfer solution from beaker into 250 ml volumetric flask. Rinse the beaker 2-3 times with distilled water and transfer this to volumetric flask. Then dilute the solution with the help of distilled water up to the mark. Resultant solution will be 0.350 M oxalic acid solution.

13) Here we have to prepare dilute solution of HCl. We can use , dilution formula to calculate volume of stock HCl solution required to prepare 0.1 M HCl solution.

We have dilution formula, M _stock $\times$ V _stock = M _dilute$\times$ V _dilute

$\therefore$ V _stock = M _dilute$\times$ V _dilute / M _stock

$\therefore$ V _stock = 0.1 M $\times$ 500.0 ml / 3.0 M = 16.66 ml = 16.7 ml

Procedure : With the help of graduated pipette , pipette out 16.7 ml of 3.0 M HCl solution & transfer it into 500 ml volumetric flask. Then add distilled water up to the mark . The resultant solution will be 0.1 M HCl solution.