Although most of that of the alum you collected in the filtration process was re

Question

Although most of that of the alum you collected in the filtration process was recovered on the filter paper, some amount of alum remained dissolved, passed through the filter paper and is now lost.. The amount of alum that can be dissolved in water to make a 50 mL solution is known 1.4 g of alum @ 1.0 degree C 1.9 g of alum @ 3.0 degree C 2.4 g of alum @ 6.0 degree C Using MS Excel, construct a mass vs. temperature scatter graph using these three points. Perform a trend line analysis and attach the graph to your report. b. Use the trendline analysis result and the temperature you recorded earlier to determine how much alum should still be dissolved in 50 mL of solution at this temperature, (show your calculations) c. Adjust the mass of dissolved alum dissolved in 50 mL (previous step) for your solution's volume ("h" in table above). d. Add the dissolved alum mass to the product weight (1) and use this value to recalculate the percent yield (Show calculations). Is the new percent yield reasonable? Why? Dividable experimental report will be due at the beginning of class next week. All entries must be in written in ink before you leave the lab. I will grade your data and calculations by entering your data table measurements into an Excel Spreadsheet, receive full credit for your calculations unless your results match mine exactly.

Explanation / Answer

From the data

moles of aluminium taken = 0.035954 mols

1 mole of Al forms 1 mole of Alum product

moles of alum product to be obtained = 0.035954 mols

Theoretical mass of alum product = 0.035954 mol x 474.3904 g/mol = 17.056 g

From the given experiment

l. mass of alum formed = 2.6 g

m. Percent yield of alum = 2.6 x 100/17.056 = 15.24%

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