I submitted this question yesterday; the first answer was right, the second was wrong. Please take another look. Thank you! 9.762 g of a non-volatile solute is dissolved in 175.0 g of water.The solute does not react with water nor dissociate in solution.Assume that the resulting solution displays ideal Raoult's law behaviour.At 25°C the vapour pressure of the solution is 23.405 torr.The vapour pressure of pure water at 25°C is 23.756 torr.Calculate the molar mass of the solute (g/mol). answer: 67.0 g/mol -- correct Now suppose, instead, that 9.762 g of a volatile solute is dissolved in 175.0 g of water.This solute also does not react with water nor dissociate in solution.The pure solute displays, at 25°C, a vapour pressure of 2.376 torr.Again, assume an ideal solution.If, at 25°C the vapour pressure of this solution is also 23.405 torr.Calculate the molar mass of this volatile solute. answer: 48.81 g/mol -- wrong Hint: Note that the calculation is greatly simplified if you recognize that: Xsolute= 1 - Xwater.Use this fact and the Raoult's Law expression for the total vapour pressure to solve for Xwater. Then express Xwater in terms of the relative masses, and solve for the molar mass of the solute. Actually, if you care to do the algebraic manipulations, a relatively simple expression can be derived for the molar mass in terms of the given masses and pressures.
I submitted this question yesterday; the first answer was right, the second was wrong. Please take another look. Thank you! 9.762 g of a non-volatile solute is dissolved in 175.0 g of water.The solute does not react with water nor dissociate in solution.Assume that the resulting solution displays ideal Raoult's law behaviour.At 25°C the vapour pressure of the solution is 23.405 torr.The vapour pressure of pure water at 25°C is 23.756 torr.Calculate the molar mass of the solute (g/mol). answer: 67.0 g/mol -- correct Now suppose, instead, that 9.762 g of a volatile solute is dissolved in 175.0 g of water.This solute also does not react with water nor dissociate in solution.The pure solute displays, at 25°C, a vapour pressure of 2.376 torr.Again, assume an ideal solution.If, at 25°C the vapour pressure of this solution is also 23.405 torr.Calculate the molar mass of this volatile solute. answer: 48.81 g/mol -- wrong Hint: Note that the calculation is greatly simplified if you recognize that: Xsolute= 1 - Xwater.Use this fact and the Raoult's Law expression for the total vapour pressure to solve for Xwater. Then express Xwater in terms of the relative masses, and solve for the molar mass of the solute. Actually, if you care to do the algebraic manipulations, a relatively simple expression can be derived for the molar mass in terms of the given masses and pressures.
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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I submitted this question yesterday; the first answer was right, the second was wrong. Please take another look. Thank you!
-
9.762 g of a non-volatile solute is dissolved in 175.0 g of water.
The solute does not react with water nor dissociate in solution.
Assume that the resulting solution displays ideal Raoult's law behaviour.
At 25°C the vapour pressure of the solution is 23.405 torr.
The vapour pressure of pure water at 25°C is 23.756 torr.
Calculate the molar mass of the solute (g/mol). answer: 67.0 g/mol -- correct - Now suppose, instead, that 9.762 g of a volatile solute is dissolved in 175.0 g of water.
This solute also does not react with water nor dissociate in solution.
The pure solute displays, at 25°C, a vapour pressure of 2.376 torr.
Again, assume an ideal solution.
If, at 25°C the vapour pressure of this solution is also 23.405 torr.
Calculate the molar mass of this volatile solute. answer: 48.81 g/mol -- wrong
Hint:
Note that the calculation is greatly simplified if you recognize that: Xsolute= 1 - Xwater. Use this fact and the Raoult's Law expression for the total vapour pressure to solve for Xwater. Then express Xwater in terms of the relative masses, and solve for the molar mass of the solute. Actually, if you care to do the algebraic manipulations, a relatively simple expression can be derived for the molar mass in terms of the given masses and pressures. |
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