Exercise 4: Commercial acids and bases are aqueous solution with following properties,Calculate the Molarity and Molality of each reagent:HintsSubstance % Mass Density (g/ml) Molar Mass Molarity MolalityHC138HNO370H₂SO4CH3COOHNH3(a) Molarity=(b) Molality=9599281.191.421.841.050.90Moles soluteVolume solution (L)Moles solutemass solvent (Kg)(c) Mass solvent = mass solution-mass solute3663986017

Molarity : It is defined as the number of moles of solute dissolved per litre of the solution .…

Exercise 4: Commercial acids and bases are aqueous solution with following properties, Calculate the Molarity and Molality of each reagent: Hints Substance HCI HNO3 H₂SO4 CH3COOH NH3 (a) Molarity (b) Molality= % Mass Density (g/ml) Molar Mass Molarity Molality 38 70 95 99 28 1.19 1.42 1.84 1.05 0.90 Moles solute Volume solution (L) Moles solute mass solvent (Kg) (c) Mass solvent = mass solution-mass solute 36 63 98 60 17

Exercise 4: Commercial acids and bases are aqueous solution with following properties, Calculate the Molarity and Molality of each reagent: Hints Substance HCI HNO3 H₂SO4 CH3COOH NH3 (a) Molarity (b) Molality= % Mass Density (g/ml) Molar Mass Molarity Molality 38 70 95 99 28 1.19 1.42 1.84 1.05 0.90 Moles solute Volume solution (L) Moles solute mass solvent (Kg) (c) Mass solvent = mass solution-mass solute 36 63 98 60 17

Introductory Chemistry: A Foundation

9th Edition

ISBN:9781337399425

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Donald J. DeCoste

Chapter15: Solutions

Section15.3: Solution Composition: Mass Percent

Problem 15.2SC

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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
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