Grams of solute needed to lower vapor pressure of solvent?

[Elvis 123456789]

Homework Statement

Some KCl is dissolved in water at 25°C, where it completely dissociates. The vapor pressure of pure water at 25°C is 28.3 mmHg. Estimate the mass in grams of KCl needed per liter of pure water to reduce the vapor pressure of water at 25°C by 5%.

Homework Equations

The Attempt at a Solution

I'm really not sure how to go about this problem. First I thought I could maybe use raoult's law, but then I don't think that would work to get the mass of the solute. Then I thought maybe if the vapor pressure is lowered by 5%, then the solvent is also lowered by 5%, which would mean that for 1 L of water I would need 0.05 L of KCl to lower the vapor pressure by 5%, but I don't think this is right either.

 

[Raghav Gupta]

Use the equation (p-p°)/p° = x where p is vapour pressure of solution after kcl is added,p° V.P of pure water, and x is mole fraction of kcl in water.
The above equation's derivation comes from raoult' law which you can google.
Now you can substitute the values.

 

[Elvis 123456789]

so 5% of 28.3mmHg would be 26.885mmHg, then 26.885mmHg= X(mole fraction) * 28.3mmHg------> X=0.95

1L H2O * (1000mL/1L) *(1g/1mL) * 1mol/18.02g = 55.49 mol H2O...mol KCl= Z

(55.49 mol H2O)/(55.49mol H2O + Z) = 0.95

Z= (55.49mol)/(0.95) - 55.49mol = 2.92 mol KCl * 74.55g/mol = 218g KCl

is this right?

 

[Raghav Gupta]

What's the molecular weight of kcl?

 

[Elvis 123456789]

74.55g/mol

 

[Raghav Gupta]

so 5% of 28.3mmHg would be 26.885mmHg, then 26.885mmHg= X(mole fraction) * 28.3mmHg------> X=0.95

1L H2O * (1000mL/1L) *(1g/1mL) * 1mol/18.02g = 55.49 mol H2O...mol KCl= Z

(55.49 mol H2O)/(55.49mol H2O + Z) = 0.95

Z= (55.49mol)/(0.95) - 55.49mol = 2.92 mol KCl * 74.55g/mol = 218g KCl

is this right?

You are wrong.
The answer coming is 207 g.
I got x=0.05.

 

[Raghav Gupta]

Why are you not applying the relative lowering of vapour pressure formula,
And directly the Raoult's law?
Do you not know that formula in post #2?

 

[Elvis 123456789]

No I am not familiar with the formula from the second post. Why can't I directly use the Raoult's law? Doesn't it make sense the way I am doing it?

 

[Raghav Gupta]

It is a bit of manipulation work.
See this
http://www.chem.purdue.edu/gchelp/howtosolveit/Solutions/vp_lowering.html
After you understand this or you know tell me.

 

[Elvis 123456789]

I look up the vapor pressure lowering formula you mentioned, and the mole fraction in that formula belongs to the solute. In the Raoult law, the mole fraction belongs to the solvent. If I use the vapor pressure lowering formula i would get z/(z+55.49 mol H2O)=0.05, so I think it is correct to use the Raoult law

 

[Raghav Gupta]

Actually vapor pressure lowering formula is a manipulation of Raoult's law. So the main credits to Raoult's law. Now I tell you the whole story-
If we put solute in a pure solvent
P1=x_solvent*P1°
here P1 is vapour pressure of solution and P1° the V.P of pure solvent.
Now when we add solute there is lowering in V.P of pure solvent.
$\Delta$P1=P1°- P1= P1°- P1°*x_solvent = P1°(1- x_solvent)
Also x_solute = 1- x_solvent
So,
$\Delta$P1= x_solute*P1°
Now
$\frac{ \Delta P1}{P1°}$ = $\frac{ P1°- P1}{P1°}$ = x_solute= $\frac{ n_ {solute}}{n_{solvent}+ n_{solute}}$ Since n_solute<< n_solvent
So
$\frac{ P1°- P1}{P1°}$ = $\frac{ n_ {solute}}{n_{solvent}}$
or more simply
$\frac{ P1°- P1}{P1°}$ = $\frac{ M_ {solvent}*w_{solute}}{M{solute}*w_{solvent}}$

 

[Borek]

It is about using Raoult's law AND the fact that sum of molar fractions of all substances present in the mixture equals 1.

Raghav: while your approach is correct, you seem to be making it unnecessarily complicated by stating "use the formula". It hides the logic behind. See paragraph above.

As far as I can tell you are both ignoring the fact KCl is dissociated.

 

[Raghav Gupta]

Yeah, early I was saying to use the formula, but after that I have also given proof of it by logic. Now what logic I am hiding?

We are ignoring the dissociation of KCl because the vant hoff factor is not given in question, otherwise that's true that observed colligative property = i * calculated colligative property.
i = 2 for KCl for complete dissociation but we don't know that it has completely dissociated.
I think OP is asking a school level topic wise problem.

 

[Raghav Gupta]

Sorry, Sorry. I have made a grave mistake in above post.
Yeah in question complete dissociation is written.
So i=2 as
KCl is in equilibrium with K⁺ + Cl^-.
Borek you are correct.
I was taking the OP in complete trouble.
That's why you are a mentor, as experience is also needed and I have few posts than you.

 

[Elvis 123456789]

I am confused now, so what is the proper way to solve the problem?

 

[Raghav Gupta]

so 5% of 28.3mmHg would be 26.885mmHg, then 26.885mmHg= X(mole fraction) * 28.3mmHg------> X=0.95

That's correct.
That X is X_solvent
Now X for solute is 1- X_solvent.
Then X_solute = n_solute/n_solvent since n_solute<< n_solvent.
From here you get mass of KCl.
But since KCl is completely dissociated.
i = 2 ( See vant hoff factor)

Actual mass of KCl = i * calculated mass

 

[Borek]

Actual mass of KCl = i * calculated mass

Nope. Think it over.

 

[Raghav Gupta]

Hmm--- So many mistakes.
So Actual concentration of KCl = i * calculated concentration of KCl

Now calculated concentration is moles of KCl/ Volume of solution in liters
Assume the volume is negligibly changed by adding KCl

 

[Borek]

So Actual concentration of KCl = i * calculated concentration of KCl

I am not convinced you got it right this time. Not sure what you mean by "actual" and "calculated" - but if the one you got from the pressure information is "calculated", it is still wrong way around.

Assume the volume is negligibly changed by adding KCl

That's not necessary, molar fractions depend only on the amount of mixed substances, volume doesn't play any role. The only assumption you need is the one about the water density (1 L = 1 kg).

 

[Raghav Gupta]

That's not necessary, molar fractions depend only on the amount of mixed substances, volume doesn't play any role.

Not got it.
I think when finding concentration in molarity
Molarity = no of moles of solute / volume of solution in liters
Since we don't know density of solute, we cannot tell what will be the volume of solution and that's why we consider volume of water only.

 

[Raghav Gupta]

I am not convinced you got it right this time. Not sure what you mean by "actual" and "calculated" - but if the one you got from the pressure information is "calculated", it is still wrong way around.

According to wikipedia-Vant hoff factor in first para.
Is that wrong, when wikipedia is a recognized site?

 

[Borek]

Wikipedia is not always right, but it is in this case. It doesn't make your calculations automatically correct, when it is not clear what you mean by "actual" and "calculated".

 

[Borek]

I think when finding concentration in molarity
Molarity = no of moles of solute / volume of solution in liters

Who told you you have to use molarity?

I have explained why it doesn't matter in the next phrase, have you read it?

 

[Raghav Gupta]

I am now going in a confusion.
Can you tell from the scratch how to go about the problem?

 

[Raghav Gupta]

Wikipedia is not always right, but it is in this case.

So as by you if Wikipedia is right in this case then in first para they have written terms actual and calculated concentration.
They are saying calculated concentration is taking not in account dissociation of KCl and actual concentration taking in account dissociation of KCl.

 

[Raghav Gupta]

We get X for solvent by using Raoult's law.

Now X for solute is 1- X_solvent.
Then X_solute = n_solute/n_solvent since n_solute<< n_solvent.
From here you get mass of KCl.
But since KCl is completely dissociated.
i = 2 ( See vant hoff factor)

Isn't this correct?

 

[Borek]

Some KCl is dissolved in water at 25°C, where it completely dissociates. The vapor pressure of pure water at 25°C is 28.3 mmHg. Estimate the mass in grams of KCl needed per liter of pure water to reduce the vapor pressure of water at 25°C by 5%.

Raoult's law:

$$p_i = x_i p^0_i$$

We deal with water. To reduce the pressure by 5% we need to lower the water mole fraction in the solution to 95%. That means the other substance must be 5% or 0.05 of the mixture (again, mole fraction).

Liter of water is 1000 g, or 55.51 moles.

We need to add such n moles of solute that

$$\frac n {n + 55.51} = 0.05$$

However, KCl dissociates, which means when we add one mole of KCl, we actually add two moles of ions. That in turn means when we replace n with n_KCl (number of moles of this particular substance) it counts twice, and the equation takes form

$$\frac {2n_{KCl}} {2n_{KCl} + 55.51} = 0.05$$

Solve for n_KCl, multiply by molar mass.

I got 108 g.

 

[Raghav Gupta]

Your method is truly simple and great but then also I have a doubt for the answer.
Can you @Elvis 123456789 confirm the answer and are you following posts?

 

[Elvis 123456789]

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?

 

[Elvis 123456789]

I also found this key for an exam and the problem is solved the way i had originally done it? ...http://rogersal.people.cofc.edu/Fire Alarm Test.pdf

 

[Raghav Gupta]

I also found this key for an exam and the problem is solved the way i had originally done it? ...http://rogersal.people.cofc.edu/Fire Alarm Test.pdf

Okay that's also correct by using only mole fraction of solvent.
But the link is ignoring the fact of complete dissociation of KCl.

 

[Raghav Gupta]

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?

The answer you are getting matches with Borek and here you are taking in account complete dissociation of KCl.
But why the link is not showing that. It is giving the answer 221g?

 

[Elvis 123456789]

Yah I don't know what's up with that. The key was apparently made by someone with a Phd in chemistry or a closely related field.

 

[Raghav Gupta]

Anyways I realize that you are correct.

 

[Elvis 123456789]

Well this has been a pretty lengthy thread. Thank you for staying with me and helping me out, its been much appreciated.