Suppose 0.163 g of lead(II) acetate is dissolved in 300. mL of a 41.0 m M aqueous solution of ammonium sulfate.Calculate the final molarity of lead(II) cation in the solution. You can assume the volume of the solution doesn't change when the lead(II) acetate is dissolved init.Be sure your answer has the correct number of significant digits.ОмM105

Molarity=molesofsolute÷volumeofsolution(L)Moles=mass÷molarmassGiven:Mass of Pb(CH3COOH)2=0.163…

Answered: Suppose 0.163 g of lead(II) acetate is…

Introductory Chemistry: A Foundation

9th Edition

ISBN:9781337399425

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Donald J. DeCoste

Chapter15: Solutions

Section: Chapter Questions

Problem 38CR

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A soft drink contains an unknown mass of citric acid, C3H5O(COOH)3. It requires 6.42 mL of 9.580 × 10−2-M NaOH to neutralize the citric acid in 10.0 mL of the soft drink. C3H5O(COOH)3(aq) + 3 NaOH(aq) → Na3C3H5O(COO)3(aq) + 3 H2O(ℓ) Determine which step in these calculations for the mass of citric acid in 1 mL soft drink is incorrect? Why? n (NaOH) = (6.42 mL)(1L/1000 mL)(9.580 × 10−2 mol/L) n (citric acid) = (6.15 × 10−4 mol NaOH) × (3 mol citric acid/1 mol NaOH) m (citric acid in sample) = (1.85 × 10−3 mol citric acid) × (192.12 g/mol citric acid) m (citric acid in 1 mL soft drink) = (0.354 g citric acid)/(10 mL soft drink) Determine the correct result.
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Suppose 13.0g of potassium sulfate is dissolved in 250.mL of a 0.20 M aqueous solution of sodium chromate. Calculate the final molarity of sulfate anion in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Be sure your answer has the correct number of significant digits.
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