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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: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter10: Solutions
Section: Chapter Questions
Problem 68QAP: Consider two solutions at a certain temperature. Solution X has a nonelectrolyte as a solute and an...
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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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