A solution was prepared by dissolving 92.1 g of iodine, 12, in 800.0 g of chloroform, CHCI3.1. Calculate the mole fraction of iodine in the solution.2. Find the boiling point of the solution. (Hint iodine is nonvolatile and non-dissociating solid) Potentially useful or useless data:Boiling point of pure chloroform = 61.2 °CMolar mass of pure chloroform = 119.4 g/molBoiling point elevation constant (Kp) for chloroform = 3.63°C-m1Molar mass of lodine = 253.8 g/molmoles of solutemoles of solutionFormula : mole fraction%3DATh =Kb mAlso, Tâ = T°b + ATBO Mol fraction = 0.103Boiling Point = 64.8°CO Mol fraction = 0.363Boiling Point = 62.8°CO Mol fraction = 0.363Boiling Point = 64.8°CO Mol fraction = 0.103Boiling Point = 62.8°C

NOTE: please note, for mole fraction values none of the options given is correct.I have solved and…

A solution was prepared by dissolving 92.1 g of iodine, 12, in 800.0 g of chloroform, CHCI3. 1. Calculate the mole fraction of iodine in the solution. 2. Find the boiling point of the solution. (Hint: iodine is nonvolatile and non-dissociating solid)

A solution was prepared by dissolving 92.1 g of iodine, 12, in 800.0 g of chloroform, CHCI3. 1. Calculate the mole fraction of iodine in the solution. 2. Find the boiling point of the solution. (Hint: iodine is nonvolatile and non-dissociating solid)

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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Question

A solution was prepared by dissolving 92.1 g of iodine, 12, in 800.0 g of chloroform, CHCI3.
1. Calculate the mole fraction of iodine in the solution.
2. Find the boiling point of the solution. (Hint iodine is nonvolatile and non-dissociating solid)

Potentially useful or useless data:
Boiling point of pure chloroform = 61.2 °C
Molar mass of pure chloroform = 119.4 g/mol
Boiling point elevation constant (Kp) for chloroform = 3.63°C-m1
Molar mass of lodine = 253.8 g/mol
moles of solute
moles of solution
Formula : mole fraction
%3D
ATh =Kb m
Also, Tâ = T°b + ATB
O Mol fraction = 0.103
Boiling Point = 64.8°C
O Mol fraction = 0.363
Boiling Point = 62.8°C
O Mol fraction = 0.363
Boiling Point = 64.8°C
O Mol fraction = 0.103
Boiling Point = 62.8°C

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