Visited[Review Topics][References]Use the References to access important values if needed for this question.The molality of ethanol, CH3CH₂OH, in an aqueous solution is 5.98 mol/kg.What is the mole fraction of ethanol in the solution?Submit AnswerRetry Entire Group 9 more group attempts remaining ptsptsptspts 2reqMpts Mpts 2reqpts 2reqpts 2reqptsMpts 2reqptspts 2req[Review Topics][References]Use the References to access important values if needed for this question.The molality of hydrochloric acid, HCI, in an aqueous solution is 5.42 mol/kg.What is the mole fraction of hydrochloric acid in the solution?Submit AnswerRetry Entire Group 9 more group attempts remaining

The molality of ethanol, CH3CH2OH, in an aqueous solution is 5.98 mol/kg.What is the mole fraction…

Answered: The molality of ethanol, CH3CH OH, in…

The molality of ethanol, CH3CH OH, in an aqueous solution is 5.98 mol/kg. What is the mole fraction of ethanol in the solution? Submit Answer Retry Entire Group 9 more group attempts remaining

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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If 0.25 mol of CuCl2*2H2O is added to make 500mL of total solution, what is that solution's molarity? Show work
Apply algebraic properties, formulas, and relationships to perform operations on real-world problems (e.g., solve for: molality. • Molality is the number of moles of solute dissolved in a kilogram of solvent. molality = moles of solute or molal, m kg of solvent Sample Problem #1: What is the molality of the example sucrose solution? First Step: Known and Unknown Quantities known unknown mass of sucrose = 0.030464 g C12H2201 molality C12H2201 = ? m mass of water = 0.981256 g H,0 Second Step: Formula molality of C12H22O11 = mol of C12H22O1 1 kg of H,0 Third Step: Rearranging, intermediate calculations, and unit conversions No rearranging required. Intermediate calculations: molar mass of C12H2201 = (12 x 12.011 g) + (22 x 1.0079 g) + (11 x 15.9994 g) = 342.299 g # mol C12H2201 = 0.030464 g C12H,O1 x 1 mol C,,H„01 = 0.000089 mol C12H22O1 1 342.299 g C12H2201 Unit conversions: 0.981256 g H,Ox_1.00000 kg_ = 0.000981256 kg H,0 1 1000.00 g Fourth Step: DELL 「田

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What is the molality of lithium ions in an aqueous solution made up of 0.302 m Lİ:PO4 assuming the compound dissociates completely? STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) %3D G 4 115.79 0.001 0.302 35.0 1 6.022 x 1023 0.906 1000 1.21 0.101 g Li:PO4 mol LisPO4 m Li:PO4 g Lit m Lit kg solvent g solvent mol Lit
Calculate the molality of a solution that is prepared by mixing 25.5 mL of CH;CH2OH (D = 0.789 g/mL) and 125.8 ml of CH3CH>CH,OH (D = 0.811 g/mL). Edit View Insert Format Tools Table
What conclusions can you draw about the solubility of gases as temperature increases? Figure 3 Figure 1 Figure 2 10 NO CH, methane 0, CO CH N2 2.0 2.0 axygen CO carbon monoxide 1.0 nitrogen He helium 20 30 Temperature ("C) 10 40 50 10 20 30 40 10 20 30 Temperature (C) Temperature ("C) Look at axes for the graphs above and point out the similarities and the differences between them. Is the trend for solubility of gases still the same? Use the graph below to explain the behavior of noble gases in water. Why does the solubility show the trend it does in the figure below? 2.0E-04 1.5E-04 Xenon 1.0E-04 Krypton 5.0E-05 Argon Neon Hellum. 0.0E+00 10 20 30 Temperature (C) Solubility (molelmole, 1 atm) Solubility (10-3 mol L-) Solubility (mg/100g H,0) ||||| Solubility (mM)
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Submit Answer If 22.2 grams of an aqueous solution of calcium acetate, Ca(CH3COO)2, contains 4.73 grams of calcium acetate, what is the percentage by mass of calcium acetate in the solution? % Ca(CH3COO)2 282 F4 % 5 T G F5 B Retry Entire Group Cengage Learning Cengage Technical Support Y H [Review Topics) [References] Use the References to access important values if needed for this question. S & 6 i 7 MacBook Air F6 N 9 more group attempts remaining U F7 J 8 M I DII K ( 9 DD F9 O L ) O A F10 P : ; command option 4 F11 { [ ? 1 Previous + 1 Save and Exit 4) Next> F12 } ] S Relaunch to update delete + return shift Ⓒ
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How many grams of NazcO3 should be dissolved in 200.0g of water to prepare a solution which is 12.5 % Na,CO, by mass? O 21.43 g O 25.00 g O 28.57 g O 27.32 g O 42.86 g Next page us activity Jump to.
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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