Vapor Pressure of Benzene at 50°C: 0.028 mmHg

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Discussion Overview

The discussion revolves around calculating the vapor pressure of benzene at 50 degrees Celsius, using the ideal gas law and the density of benzene vapor. Participants explore the application of relevant equations and clarify their understanding of vapor pressure in the context of a homework problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant attempts to use the ideal gas law (PV=nRt) to find the vapor pressure of benzene, substituting density into the equation.
  • Another participant expresses confusion about the calculations and the correct application of the equations, seeking clarification on their approach.
  • Participants discuss the molar mass of benzene, with one noting a correction in their earlier calculation of it.
  • Multiple participants share the same initial misunderstanding regarding the calculation process and the units for the answer.

Areas of Agreement / Disagreement

There is no consensus on the initial calculation method, as participants express confusion and seek help. However, one participant later claims to have arrived at the correct answer after correcting their molar mass calculation.

Contextual Notes

Participants mention the lack of units provided in the homework problem, which contributes to their uncertainty in determining the correct answer.

Who May Find This Useful

Students studying vapor pressure calculations, particularly in the context of ideal gas behavior and homework problem-solving in chemistry.

ally1h
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Homework Statement


Benzene, C6H6, has a boiling point of 80 degrees Celsius. The density of benzene vapor in equilibrium with liquid benzene at 50 degrees Celsius is 1.153 g/L. What is the vapor pressure of benzene in mmHg at 50 degrees Celsius?


Homework Equations


PV=nRt
density = m/v



The Attempt at a Solution


I thought PV=nRt might cover it.. substituting d=mv into the equation and getting:
P = dRt/m
Plugging in I get:
[(1.153 g/L)(62.364 L Torr/mol K)(323 K)] / (1084.1 g/mol)
P = 21.43 Torr = 0.028 mmHg


What am I doing wrong? My professor covered what vaporization is, not how to go about getting vaporization pressure. I have a test Wednesday and I don't know how to do any of it!
 
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ally1h said:

Homework Statement


Benzene, C6H6, has a boiling point of 80 degrees Celsius. The density of benzene vapor in equilibrium with liquid benzene at 50 degrees Celsius is 1.153 g/L. What is the vapor pressure of benzene in mmHg at 50 degrees Celsius?


Homework Equations


PV=nRt
density = m/v



The Attempt at a Solution


I thought PV=nRt might cover it.. substituting d=mv into the equation and getting:
P = dRt/m
Plugging in I get:
[(1.153 g/L)(62.364 L Torr/mol K)(323 K)] / (1084.1 g/mol)
P = 21.43 Torr = 0.028 mmHg


What am I doing wrong? My professor covered what vaporization is, not how to go about getting vaporization pressure. I have a test Wednesday and I don't know how to do any of it!

d = m / v
 
Sorry, that was a typo. I did divide when doing the actual problem. But I'm still not getting the answer. I forgot to add the choices on our homework handout.

a) 298
b) 50
c) 352
d) 462
e) 0.392


I still don't understand what I am doing incorrectly. Please help?
 
ally1h said:
Sorry, that was a typo. I did divide when doing the actual problem. But I'm still not getting the answer. I forgot to add the choices on our homework handout.

a) 298
b) 50
c) 352
d) 462
e) 0.392


I still don't understand what I am doing incorrectly. Please help?





What are the units in the answers?
 
What is molar mass of benzene?
 
The homework problem didn't give units for the answer so I have no clue. So I have no clue as to what to aim for. The molar mass of Benzene is 78.108 g/mol.
 
Oh! Wow, duh, I feel stupid. I calculated the molar mass incorrectly. Haha, I've got the correct answer now! It's choice A, 298!

haha, thanks for pointing out my mistake!
 

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