9.294 g of a non-volatile solute is dissolved in 305.0 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 60°C the vapour pressure of the solution is 148.07 torr.The vapour pressure of pure water at 60°C is 149.40 torr.Calculate the molar mass of the solute (g/mol).61.1 g/molYou are correct.Your receipt no. is 162-1722 ?Previous TriesNow suppose, instead, that 9.294 g of a volatile solute is dissolved in 305.0 g of water.This solute also does not react with water nor dissociate in solution.The pure solute displays, at 60°C, a vapour pressure of 14.94 torr.Again, assume an ideal solution.If, at 60°C the vapour pressure of this solution is also 148.07 torr.Calculate the molar mass of this volatile solute.Note that the calculation is greatly simplified if you recognize that: \solute = 1- Xwater.

Step 1: For an ideal solution containing 2 volatile components (in this case, water and solute), the…

Answered: 9.294 g of a non-volatile solute is…

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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An experiment conducted at 20 oC requires an atmosphere in which the partial pressure of water is a constant 10.0 torr. The experiment will be conducted inside a sealed container in which there is a beaker containing a solution of ethylene glycol (FW 62.07) and water (FW 18.02). At 20 oC, the vapor pressure of pure ethylene glycol is 0.056 torr and that of pure water is 17.3 torr. What mass of ethylene glycol must be mixed with 200. g water to produce a solution with a 10.0 torr vapor pressure of water? A separate solution is prepared by mixing 5.00 g ethylene glycol and 200. g water. Calculate the molality of ethylene glycol in the solution.
Osmosis is the process responsible for carrying nutrients and water from groundwater supplies to the upper parts of trees. Since the tree sap has a higher solute concentration than ground water, water passes through root membranes into the tree, where osmotic pressure can be as high as 20 atm in the tallest trees.What would the molar concentration of the tree sap (a non-electrolyte) have to be to achieve an osmotic pressure of 17.9 bar on a day when the temperature is 22.5 °C?
2. A and B form an ideal solution. At a total pressure of 0.90 bar, Xa(1)=0.65, XA(g) = 0.45. Using this information, calculate the vapor pressures of pure A and pure B.
. The vapor pressure of benzene is 10 kPa while methylbenzene is 2.8 kPa. Determine the total pressure of an equimolar solution.
33. 1.18 g aniline liquid is mixed with 1.00 g benzene liquid at 25 oC. Calculate the mole fraction of aniline in the vapor phase. At 25 oC, the vapor pressure of pure benzene is 30.5 kPa and the vapor pressure of pure aniline is 28.0 kPa. molar masses: benzene = 78.00 g/mole and aniline = 93.12 g/mole Group of answer choices 0.522 0.233 0.821 0.476
A certain dilute solution has an osmotic pressure of 11.1 atm at 25°C. Calculate the difference between the chemical potential of the solvent in the solution and that of pure water. Assume that the density is the same as that of water at 25°C, 0.99705 g-cm-3. (Hint: Express the chemical potential in terms of mole fraction, x1, and rewrite the osmotic pressure equation as OV = n2RT, where n2 is the number of moles of the solute and V = 1.00 L.)
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…
A salt solution has an osmotic pressure of 17.0 atmospheres at 17°C. What assumptions must be made to calculate the freezing point of this solution? molality and molarity of the solution are approximately equal and that the electrolyte is behaving ideally molarity is twice the molality and the electrolyte (salt) is behaving ideally molality and molarity of the solution are approximately equal and there is no electrolyte any more salt is KCl the freezing point is at 0°C
An experiment conducted at 20 oC requires an atmosphere in which the partial pressure of water is a constant 10.0 torr. The experiment will be conducted inside a sealed container in which there is a beaker containing a solution of ethylene glycol (FW 62.07) and water (FW 18.02). At 20 oC, the vapor pressure of pure ethylene glycol is 0.056 torr and that of pure water is 17.3 torr. What mass of ethylene glycol must be mixed with 200. g water to produce a solution with a 10.0 torr vapor pressure of water? (Report the answer to three significant figures.)

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