Exercise 1(a) A 10-g quantity of a compound is dissolved in 100 g of water. The solution formed has a meltingpoint of -7.14°C. Is the compound KCI, NazS, or Na3PO4?(b) An aqueous solution of a nonelectrolyte is made by dissolving the solute in 1 L of water. Thesolution has a freezing point of -1.5°C. What volume of water must be added to change thefreezing point to -1.15°C?(c) 100 mL of a solution of physiological saline (0.92% NaCl by mass) is diluted by the addition of250 mL of water. What is the osmotic pressure of the final solution at 37°C? Assume that NaCldissociates completely into Na*and Cl (aa).(aa)

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Answered: Exercise 1 (a) A 10-g quantity of a…

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

Section: Chapter Questions

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
QUESTION 1 An aqueous solution containing 72.9 g of an unknown substance in 611 g water was found to have a freezing point of -1.69 °C. Calculate the molar mass of the unknown substance (Kf for water is 1.86 °C/m)
What is the vapor pressure of a solution made by dissolving 28.5 grams if C3H3O3 in 125 grams of water at 30°c? The vapor pressure of pure water at that temperature is 31.8 mm Hg. « < Questio response.
1st attempt l See Periodic Table O See Hint A solution prepared by dissolving 513 mg of a sugar (a molecular compound and a nonelectrolyte) in 1.00 g of water froze at -2.79°C. What is the molar mass of this sugar? The value of Kf is 1.86°C/m. g/mol
The presence of what ions make water hard? (select all that apply) Na+ CO3²- Mg2+ Fe2+ CI™ OCa²+
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Exercise 4: Commercial acids and bases are aqueous solution with following properties, Calculate the Molarity and Molality of each reagent: Hints Substance % Mass Density (g/ml) Molar Mass Molarity Molality HC1 38 HNO3 70 H₂SO4 CH3COOH NH3 (a) Molarity= (b) Molality= 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
What would be the freezing point, in degree Celsius, of a 3.8 m aqueous solution of Sc(CIO4)3? kç = 1.86 °C/m 28.3 X 7.07 Your answer -28.3 Correct answer D 7.07 AI
QUESTION 23 What is the freezing point (°C) of a solution prepared by dissolving 6.10 g of (NH4)2SO4 (formula weight = 132 g/mol) in 135 g of water? The molal freezing point depression constant for water is 1.86°C/m. Assume complete dissociation in solution. Express your answer in numerical form (NOT scientific notation) to 3 significant figures.
Question 3 Water has a Kf of -1.86 °C/m. Calculate the new freezing point of an aqueous solution made by mixing 28.85 g water and 1.84 g MgCl2. Submit Question
QUESTION 31 What would be the theoretical freezing point of a solution made by dissolving 10.03 g of Crso in 240 0 g of water? The treezing point constant of water is 1.86 "Cimola O 0.9919 C -1.000 °C -0.9919 "C O -1.008 "C O 1.008 "C QUESTION 32 If the dissolution of a salt in water is an endothermic reaction, how will the temperature of the solution change as the reaction proceeds? O There is insuttucient information to answer the question O The solution will cool down. O There wil be no change in temperature O The solution will heat up

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