Given that I used 0.404 g Sample 1 and 0.403 g Sample 2 to begin with. Also, Cru

Question

Given that I used 0.404 g Sample 1 and 0.403 g Sample 2 to begin with. Also, Cruicible weight for Sample 1: 30.678 g and Crucible weight for Sample 2: 30.069 g.
For the paper version, I got 31.056 g (Precipitate + Crucible wt) for Sample 1, which after calculations was 23.14% Cl present.
For the online table version I typed, I added an extra “1 g” to the Precipitate + Crucible wt for Sample 1, which after calculations was 84.4% Cl present
SOLUBLE CHLORIDE IN A SAMPLE Record the Sample ID in your notebook. You will be reporting your result with this ID number. Weigh out 2 samples of approximately 0.4 g each (you must know exact weight nearest to 0.1 mg) in two separate 250 ml beakers. Dissolve each sample in 100 ml of 0.1 M HNO3. Keeping out from bright light, precipitate silver chloride by adding about 35 ml of 0.20 M AgNO3. Digest the solution at low heat over a hot plate. When the precipitates separate out, add a few drops of AgNOs to the clear supernatant to check if the precipitation is complete. If fresh precipitates develops, add additional 10 ml of AgNO3. Cl (aq) + AgNO3 (aq)AgCI (s) Let the solution stand overnight for complete precipitation. Weigh and number two filter crucibles and filter the precipitates through the crucibles using suction. Transfer quantitatively all precipitates into the crucible by rinsing the beaker with deionized water Wash precipitates with O.1 M HNO3 until all excess AgNO3 is removed. Check wash solution with dilute HCIl. If no precipitates develops, washing is complete. Dry the crucible in an oven to constant weight. Final precipitate should be white or little pink. Subtract the weight of filter crucible and use the appropriate gravimetric factor to calculate the amount of chloride in the sample. ravimetric factor for Clin AgCl = Atomic wt. of Cl / Formula wt. AgCl

Explanation / Answer

1 mol AgCl has 1 mol of Cl atoms

or ,1mol of AgCl(=143.32 g/mol) has 1 mol of Cl(=35.45g/mol)

So % by mass of Cl in 1 mol AgCl =(35.45/143.32)*100=24.73%

As % Cl is close to that obtained from calculations in your paper version,so correct value is from the experimentally obtained mass of precipitate (paper version)

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