Consider the given liquid-vapor diagram: 375 370 365 360 355 350 345 340 0.1 02 0.8 03 04 05 0.5 0.7 09 A A Mole fraction B What will be the composition of the vapor when a mixture containing 0.30 of A was subjected to fractional distillation with two theoretical plates? O 0.72 A: 0.28 B 0.87 A: 0.13 B 0.50 A: 0.50 B Temperature (K)

Consider the given liquid-vapor diagram: 375 370 365 360 355 350 345 340 0.1 02 0.8 03 04 05 0.5 0.7 09 A A Mole fraction B What will be the composition of the vapor when a mixture containing 0.30 of A was subjected to fractional distillation with two theoretical plates? O 0.72 A: 0.28 B 0.87 A: 0.13 B 0.50 A: 0.50 B Temperature (K)

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...

See similar textbooks

Similar questions

1-Propanol (P = 20.9 Torr at 25 °C) and 2-propanol (P2 = 45.2 Torr at 25 °C) form ideal solutions in all proportions. %3D Let a1 and x2 represent the mole fractions of 1-propanol and 2-propanol in a liquid mixture, respectively, and y1 and represent the mole fractions of each in the vapor phase. Y2 For a solution of these liquids with a1 0.310, calculate the composition of the vapor phase at 25 °C.
%3D 28. K, (Alloxanic acid) = 2.3 x 107 and Ka (HF) = 6.8 x 10ª. Consider the following two aqueous mixtures: %3D Solution (1): mixing 50.0 mL each of 0.25-M Alloxanic acid and 0.25-M NaOH. Solution (2): mixing 50.0 mL each of 0.25-M HF and 0.25-M NAOH. Part 1. Which of the above mixtures does have a higher pH after the solutions reach equilibrium? b) Solution (2) c) same pH for both solutions d) insufficient information. a) Solution (1) Part 2. Which of the above two mixtures in Part 1 does have pH > 7 at equilibrium? a) Solution (1) b) Solution (2) c) both solutions d) neither of the two solutions
Boiling Point Elevation/Freezing Point Depression T = m K where, for freezing point depression: T = T(pure solvent) - T(solution) and for boiling point elevation: T = T(solution) - T(pure solvent) m = (# moles solute / Kg solvent) Kb = boiling point elevation constant. Kf = freezing point depression constant. Kb and Kf depend only on the SOLVENT. Below are some common values. Use these values for the calculations that follow. Solvent Formula Kb(°C / m) Kf(°C / m) Water H2O 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHCl3 3.67 Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02
12 g of unknown organic sample was dissolve in 706 mL of Dicloromethane (DCM). The boiling point of benzene was increased by 2.72oC. Determine the molecular weight of the unknown sample? Kb of DCM = 2.42oC/m Bb of benzene = 39.6 oC density of benzene = 1.33 g/mL at 25 °C Round your answer to the nearest whole number, no units required.
Quinine is a natural product extracted from the bark of the cinchona tree, which is native to South America. Quinine is used as an antimalarial agent. When 1.90 g of quinine is dissolved in 25.0 g of cyclohexane, the freezing point of the solution is lowered by 4.88 °C. Look up the freezing point and Kf constant for cyclohexane in the Colligative Constants table. Calculate the molar mass of quinine. molar mass: g/mol
Answer number 1 with complete solution.
Colligative Constants Constants for freezing-point depression and boiling-point elevation calculations at 1 atm: Kvalue Normal freezing (°C/ m) point (°C) water H₂0 1.86 benzene C6H6 5.12 cyclohexane C6H 12 20.8 ethanol C₂H6O 1.99 carbon tetrachloride 29.8 CCI 4 camphor C 10 H 16 0 37.8 Solvent Formula- 0.00 5.49 6.59 -117.3 -22,9 176 K, value Normal boiling (°C/ m) point (°C) 0.512 100.00 2.53 2.92 1.22 5.03 80.1 80.7 784 76.8
What is the molar ratio of 1,1-dichloroethane (11DCE) to 1,2-dichloroethane (12DCE) in the gas phase above the solution containing 10 mol of 1,1-dichloroethane and 10 mol of 1,2-dichloroethane at 20oC. The vapor pressure of 1,1-dichloroethane is 24.0kPa and the vapor pressure of 1,2-dichloroethane is 8.0 kPa at 20oC.
Hi there. Please help solve this. Thanks
17. The vapor pressures of each component in a mixture of propanone (acetone, A) and trichloromethane (chloroform, C) were measured at 35 °C with the following results: 0 0 347 0.20 35.3 250 0.40 82.5 175 0.60 142 92.3 XC Pc (torr) PA (torr) What is the Henry's law constant for chloroform in this system? A) 365 torr B) 347 torr C) 177 torr D) 273 torr E) 485 torr 0.80 200 36.8 1 273 0
If a solution containing 4.30 g triphenylmethane (C6H5)3CH, in 800 g of benzene is cooled to 0.321 °C below the freezing point of pure benzene a) how much solvent will crystallize out and b) what will be the molality of the residual solution?
A solution is prepared using 19 mL of hexane and 23 mL of heptane. What is the mole fraction of hexane in the vapor above the solution? Provide your answer to the nearest hundredth. The information below may be useful to you: MW (g/moL) density (g/mL) Standard Vapor Pressure (kPa) hexane 86.15 0.655 20.1 heptane 100.21 0.684 6.1