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Sample Exercise 5: Calculation of Mole Fraction and molality An aqueous solution of HCl contains 36% HCI by mass. Calculate: (a) the mole fraction of HCI and (b) molality of HCl in the solution.

Sample Exercise 5: Calculation of Mole Fraction and molality An aqueous solution of HCl contains 36% HCI by mass. Calculate: (a) the mole fraction of HCI and (b) molality of HCl in the solution.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Sample Exercise 5: Calculation of Mole Fraction and molality An aqueous solution of HCI contains 36% HCI by mass. Calculate: (a) the mole fraction of HCI and (b) molality of HCI in the solution. Solution (a) 36 mass % => 36 g of HCI in 100.0 g of total solution => 36 g HCl and 100.0 g -36 g = 64 g H₂O moles HCI: MMHCI = 36.453 g/mol 36 g 36.453 g/mol = 0.9876 mol Total moles || MM H₂O = 18.01 g/mol moles H₂O: 64 g 18.01 g/mol 3.5526+0.9876 = 4.5402 mol Thus, X HG = 0.9876 mol = 0.22 4.5402 mol m = moles solute kg solvent = 3.5526 mol (b) Since we know the moles of the solute and the mass of the solvent, we easily can calculate the molality: 0.9876 mol = 15 m 0.064 kg
Transcribed Image Text:Sample Exercise 5: Calculation of Mole Fraction and molality An aqueous solution of HCI contains 36% HCI by mass. Calculate: (a) the mole fraction of HCI and (b) molality of HCI in the solution. Solution (a) 36 mass % => 36 g of HCI in 100.0 g of total solution => 36 g HCl and 100.0 g -36 g = 64 g H₂O moles HCI: MMHCI = 36.453 g/mol 36 g 36.453 g/mol = 0.9876 mol Total moles || MM H₂O = 18.01 g/mol moles H₂O: 64 g 18.01 g/mol 3.5526+0.9876 = 4.5402 mol Thus, X HG = 0.9876 mol = 0.22 4.5402 mol m = moles solute kg solvent = 3.5526 mol (b) Since we know the moles of the solute and the mass of the solvent, we easily can calculate the molality: 0.9876 mol = 15 m 0.064 kg
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