Raoult's law - Vapor pressure

In summary, the conversation is discussing the calculation of the grams of C10H8 present in a solution of toluene based on its vapor pressure and density. The approach involved using Raoult's equation, the density and volume of toluene, and the mole fraction equation. The final result was 12.8g, but the book states it should be 8.03g. There may have been errors in rounding or calculation, but the individuals believe their approach is correct.
  • #1
Wek
18
0
The density of toluene is 0.862g/ml at 25C and it's vapor pressure s 28.44 mmHg. At 25C, the vapor pressure of toluene above a solution of C10H8 in 500ml toluene is 27.92 mmHg. How many grams of C10H8 are present in the solution?

I first used Raoult's equation to find X. Then I used the density and given volume of toluene to find the grams and then the moles of it. Then using the moles fraction equation I found the moles of C10H8 and using the FW of it I found the grams of it.

I did the work and come up with 12.8g, which is wrong. The book says it's 8.03g.

Can anyone tell me what I might have done wrong? Thanks
 
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  • #2
Sounds like your approach is a correct one, show your numbers.

I didn't get neither 12.8g nor 8.03g :confused: Perhaps it is too early.
 
  • #3
This is what I did:

-To find mole fraction of toluene I divided the vapor pressure of toluene solution by toluene's vapor pressure: 27.92/28.44 = 0.9817
-To get the moles of toluene: density x toluene volume then divided by toluene FW: 431g/92.0g = 4.68mol
-Using the moles fraction equation I found the moles of C10H18: C10H18 mol = (4.68/0.9817) - 4.68 = 0.10mol
-To get the mass of C10H18 I multiplied the moles of C10H18 by its FW: 0.10mol x 128g = 12.8g
 
  • #4
1. Are you rounding down intermediate results? If so - don't.

2. C10H8 or C10H18?

Wek said:
C10H18: C10H18 mol = (4.68/0.9817) - 4.68 = 0.10mol

Check your math.

You are not getting 8.03g, but (after taking into account math errors) you are very close to my result. Either we are both wrong, or the book is wrong. I tend to think we are right.
 
  • #5
for your help!

It is difficult to determine exactly where you may have gone wrong without seeing your calculations. However, it is possible that you may have used the wrong values for the vapor pressure of toluene. The given value of 28.44 mmHg is for pure toluene, but in the solution, the vapor pressure is slightly lower at 27.92 mmHg. This slight difference can affect the final calculation and result in a different value for the amount of C10H8 present in the solution. It is important to carefully check and use the correct values for all variables in the equations when solving problems in chemistry.
 

1. What is Raoult's law?

Raoult's law is a scientific law that describes the relationship between the vapor pressure of a liquid and the mole fraction of its components in a solution. It states that the partial vapor pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent multiplied by its mole fraction in the solution.

2. Who was Raoult and why is this law named after him?

François-Marie Raoult was a French chemist who first discovered and published this law in 1887. The law was named after him as a tribute for his contributions to the field of physical chemistry.

3. How is Raoult's law used in real-life applications?

Raoult's law is used in various fields such as pharmaceuticals, food and beverage production, and chemical engineering. It is used to predict the behavior of solutions and to determine the purity of solvents in a solution. It is also helpful in understanding the vapor-liquid equilibrium in distillation processes.

4. What are the assumptions made in Raoult's law?

The main assumptions of Raoult's law are that the molecules of the solute do not interact with each other or with the solvent, and that the molecules of the solvent do not interact with each other. It also assumes ideal behavior, meaning that the solvent and solute molecules are in constant motion and do not form any bonds with each other.

5. How does temperature affect Raoult's law?

The vapor pressure of a liquid generally increases with increasing temperature. Therefore, as the temperature increases, the vapor pressure of the solvent will also increase, resulting in a higher concentration of the solvent in the vapor phase. This means that the mole fraction of the solvent in the solution will decrease, and Raoult's law will no longer accurately predict the behavior of the solution.

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