molarity of BaCl2=0.2M molarity of H2SO4=6M Gravimetric and Volumetric Analysis

Question

molarity of BaCl2=0.2M
molarity of H2SO4=6M

Gravimetric and Volumetric Analysis 10.oim25.n L smL ws 1,0 Report Gravimetric and Volumetric Analysis DATA B. Section Gravimetric Determination of Sulfate Ton Trial 1 Trial 2 Trial 3 Trial 4 La15 Mass of BaSO, and filter paper, g 16 Mass of filter paper, g Mass of BaSO4 precipitate, g Millimoles of BaSO4 Millimoles of H2SO4 that reacted Volume of final solution reacted, mL [H,SO4], final solution, M H2SO4], stock solution, M* (H SO4], initial solution, M* 05 L03 631 36 Average [H SO4], initial soln, M

Explanation / Answer

Trial 1

Molar mass of BaSO4 = 233.43 g/ mol

Therefore molimoles of BaSO4 produced = 0.157 / 233.43 = 6.726 x 10-4 mols = 0.6726 mili moles

milimoles of H2SO4 that reacted = milimoles of BaSO4 produced = 0.6726 milimoles

Molality of initial/ original H2SO4 is given to be 6M

Therefore [H2SO4] in stock solution = 10 ml x 6 M/ 60 ml = 1 M

and [H2SO4] in final solution = 25 ml x 1 M/ 100 ml = 0.25 M

For Trial 2

Therefore molimoles of BaSO4 produced = 0.1556 / 233.43 = 6.666 x 10-4 mols = 0.6666 mili moles

milimoles of H2SO4 that reacted = milimoles of BaSO4 produced = 0.6666 milimoles

all other is same a trial 1

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