10.00 ml of appropriately 6M sulfuric acid is transferred to a 100-mL volumetric

Q: 10.00 ml of appropriately 6M sulfuric acid is transferred to a 100-mL volumetric

4. from dilution formula M1V1 = M2V2 10*6 = M2*100 M2 = 10*6/100 = 0.6 M 0.6*10 = M3*100 M3 = molarity of final solution = 0.06 M 5. No of mol of H2SO4 = 0.0017 mol No of mol of BaCl2 = 0.0017 mol No of mol of BaSO4 produced = 0.397 /233.43 = 0.0017 mol molarity of H2so4 = 0.0017/10*1000 = 0.17 M

ID: 958780 • Letter: 1

Question

10.00 ml of appropriately 6M sulfuric acid is transferred to a 100-mL volumetric flask and diluted to the mark with distilled water and mixed. Then 10.00 mL of this solution is diluted further to 100 mL. What is the molarity of this last solution? 10.00 mL of the final acid solution is reacted with excess barium chloride to produce a precipitate of barium sulfate (FW = 233.4 g/mol). The dry solid weighs 0.397 g. Use this mass and the dilution volumes to calculate the actual molarity of the sulfuric acid in the initial solution.

Explanation / Answer

4. from dilution formula

M1V1 = M2V2

10*6 = M2*100

M2 = 10*6/100 = 0.6 M

0.6*10 = M3*100

M3 = molarity of final solution = 0.06 M

No of mol of H2SO4 = 0.0017 mol

No of mol of BaCl2 = 0.0017 mol

No of mol of BaSO4 produced = 0.397 /233.43 = 0.0017 mol

molarity of H2so4 = 0.0017/10*1000 = 0.17 M

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10.00 ml of appropriately 6M sulfuric acid is transferred to a 100-mL volumetric

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