Grams of solute needed to lower vapor pressure of solvent?

In summary: Ok, I think I got it now. So the "calculated concentration" would be the concentration you get from using the mole fraction in the Raoult's law equation and the "actual concentration" would be the concentration you get from using the mole fraction in the vapor pressure lowering formula?Yes, that's correct. And from the "actual concentration" we can calculate the actual mass of KCl needed, taking into account the dissociation and the vant hoff factor.
  • #36
You are welcome. I hope you would do the same for somebody but in a short way.:smile:
 
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  • #37
Elvis 123456789 said:
Why is it necessary to get the mole fraction of the solute? Isn't it enough to have the mole fraction of the solvent "X=0.95"

(55.51mol H2O)/(2z + 55.51mol H2O)=0.95--------> z= 1.46mol KCl * 74.55g/mol -----> z= 109g KCl

Is this way valid? Or is it necessary to use the other formula?

Both approaches are perfectly valid and equivalent.
 
<h2>1. How is the number of grams of solute needed to lower vapor pressure of a solvent determined?</h2><p>The number of grams of solute needed to lower the vapor pressure of a solvent is determined by the Raoult's law, which states that the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent. This can be represented by the equation: Psolution = Xsolvent * P0solvent, where Psolution is the vapor pressure of the solution, Xsolvent is the mole fraction of the solvent, and P0solvent is the vapor pressure of the pure solvent.</p><h2>2. What is the relationship between the amount of solute and the lowering of vapor pressure?</h2><p>The amount of solute and the lowering of vapor pressure have an inverse relationship. This means that as the amount of solute increases, the vapor pressure of the solution decreases. This is because the solute particles take up space on the surface of the solvent, making it more difficult for the solvent molecules to escape and form vapor.</p><h2>3. How does the type of solute affect the amount of grams needed to lower vapor pressure?</h2><p>The type of solute does not have a significant effect on the amount of grams needed to lower vapor pressure. The main factor that determines the amount of solute needed is the mole fraction of the solvent. However, different solutes may have different molecular weights, which can affect the number of moles needed to achieve the desired vapor pressure lowering.</p><h2>4. What is the significance of lowering the vapor pressure of a solvent?</h2><p>Lowering the vapor pressure of a solvent is important in various industrial and laboratory processes. It can be used to control the boiling point of a solution, which is crucial in distillation and other separation techniques. It can also prevent evaporation of volatile solvents in closed systems, such as in chemical reactions or storage of liquids.</p><h2>5. Can the number of grams of solute needed to lower vapor pressure be calculated for any solvent?</h2><p>Yes, the number of grams of solute needed to lower vapor pressure can be calculated for any solvent as long as the vapor pressure and mole fraction of the solvent are known. However, it is important to note that Raoult's law is only applicable to ideal solutions, so it may not be accurate for all solvents. Additionally, other factors such as intermolecular forces and non-ideal behavior of solutes can also affect the vapor pressure of a solution.</p>

1. How is the number of grams of solute needed to lower vapor pressure of a solvent determined?

The number of grams of solute needed to lower the vapor pressure of a solvent is determined by the Raoult's law, which states that the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent. This can be represented by the equation: Psolution = Xsolvent * P0solvent, where Psolution is the vapor pressure of the solution, Xsolvent is the mole fraction of the solvent, and P0solvent is the vapor pressure of the pure solvent.

2. What is the relationship between the amount of solute and the lowering of vapor pressure?

The amount of solute and the lowering of vapor pressure have an inverse relationship. This means that as the amount of solute increases, the vapor pressure of the solution decreases. This is because the solute particles take up space on the surface of the solvent, making it more difficult for the solvent molecules to escape and form vapor.

3. How does the type of solute affect the amount of grams needed to lower vapor pressure?

The type of solute does not have a significant effect on the amount of grams needed to lower vapor pressure. The main factor that determines the amount of solute needed is the mole fraction of the solvent. However, different solutes may have different molecular weights, which can affect the number of moles needed to achieve the desired vapor pressure lowering.

4. What is the significance of lowering the vapor pressure of a solvent?

Lowering the vapor pressure of a solvent is important in various industrial and laboratory processes. It can be used to control the boiling point of a solution, which is crucial in distillation and other separation techniques. It can also prevent evaporation of volatile solvents in closed systems, such as in chemical reactions or storage of liquids.

5. Can the number of grams of solute needed to lower vapor pressure be calculated for any solvent?

Yes, the number of grams of solute needed to lower vapor pressure can be calculated for any solvent as long as the vapor pressure and mole fraction of the solvent are known. However, it is important to note that Raoult's law is only applicable to ideal solutions, so it may not be accurate for all solvents. Additionally, other factors such as intermolecular forces and non-ideal behavior of solutes can also affect the vapor pressure of a solution.

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