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Topic: Solubility of naphthalene in CCl4  (Read 3043 times)

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Offline cseil

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Solubility of naphthalene in CCl4
« on: September 04, 2015, 12:22:42 PM »
Hi everyone,
my book asks me to calculate the solubility of naphthalene (2) in CCl4 at 4°C knowing that a saturated solution of benzene (1) + naphthalene contains the 36.7% in weight of naphthalene at 21°C.

The ΔH is 4800 cal mol-1 and it is constant in the considered range of temperature. The solutions are ideal.

I think that the first thing to do is to calculate the melting temperature of naphthalene. Knowing that I want to use it to calculate the solubility in CCl4.

[tex]ln(x_2)=\frac{\Delta H}{R} (\frac{1}{T_f}-\frac{1}{T})[/tex]

The first part is to calculate Tf. I have to convert the mass fraction to the molality.
I considered 100g of solution. 36.7g of naphthalene are 0.286mol. The rest is benzene (63.3g).
So m=0.286/0.0633=4.51m.

I can convert the molality to molar fraction:
[tex]x = \frac{M_0 m}{1000} = 0.35[/tex]

Solving the equation I wrote before I calculate Tf but it's not 80°C as it should be. Why? What's wrong?
« Last Edit: September 04, 2015, 12:40:55 PM by cseil »

Offline mjc123

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Re: Solubility of naphthalene in CCl4
« Reply #1 on: September 04, 2015, 06:55:27 PM »
Your equation is wrong. Mole fraction does not vary linearly with molality. The LHS is not x but x/(1-x). (Can you see where this comes from?)
But why go through molality? You know the masses of benzene and naphthalene, and their molecular weights, so you know how many moles of each, so you can get the mole fraction directly. Why complicate things?

Offline cseil

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Re: Solubility of naphthalene in CCl4
« Reply #2 on: September 10, 2015, 10:47:50 AM »
So if I consider 100g of solution I get 0.286 mol of naphthalene and 0.810mol of benzene.

x1=0.739 (benzene)
x2=0.261

If I replace it in the first equation I cannot get the Tf of the naphthalene (I multiply the ΔH for the number of moles that I'm considering, 0.286).

EDIT: Here's my mistake. I don't know why but I multiplied dH for the moles of the 100g solution. Now I get the right answer.

So I put my Tf into the same equation to calculate the molar fraction that is soluble into CCl4 at 4°C.
I get x2=0.159.

I want to calculate the mass percentage. So I consider 1 mole of CCl4 = 153.82g, I set the equation:

[tex]x2=0.159=\frac{n2}{n2+1}[/tex]

n2= 0.189 mol = 24.23g

The weight of the solution is 153.82 + 24.23g, so w/w%=13.6% and Hallelujah!  ;D Thanks
« Last Edit: September 10, 2015, 11:01:32 AM by cseil »

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