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boyakko [2]
2 years ago
10

(4) Calculate the % of a compound that can be removed from liquid phase 1 by using ONE to FOUR extractions with a liquid phase 2

. Assume that K = 2 and the volume of phase 2 equals to 50% that of phase 1.
Chemistry
1 answer:
maksim [4K]2 years ago
5 0

Answer:

One extraction: 50%

Two extractions: 75%

Three extractions: 87.5%

Four extractions: 93.75%

Explanation:

The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.

qⁿ = (V₁/(V₁ + KV₂))ⁿ

We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:

qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ =  (V₁/(2V₁))ⁿ = (1/2)ⁿ

When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.

When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.

When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.

When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.

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a) Whatis the composition in mole fractions of a solution of benzene and toluene that has a vapor pressure of 35 torr at 20 °C?
iogann1982 [59]

Answer:

molar composition for liquid

xb= 0.24

xt=0.76

molar composition for vapor

yb=0.51

yt=0.49

Explanation:

For an ideal solution we can use the Raoult law.

Raoult law: in an ideal liquid solution, the vapor pressure for every component in the solution (partial pressure) is equal to the vapor pressure of every pure component multiple by its molar fraction.  

For toluene and benzene would be:

P_{B}=x_{B}*P_{B}^{o}

P_{T}=x_{T}*P_{T}^{o}

Where:

P_{B} is partial pressure for benzene in the liquid  

x_{B} is benzene molar fraction in the liquid  

P_{B}^{o} vapor pressure for pure benzene.  

The total pressure in the solution is:

P= P_{T}+ P_{B}

And  

1=x_{B}+x_{T}

Working on the equation for total pressure we have:

P=x_{B}*P_{B}^{o} + x_{T}*P_{T}^{o}

Since x_{T}=1-x_{B}

P=x_{B}*P_{B}^{o} + (1-x_{B})*P_{T}^{o}

We know P and both vapor pressures so we can clear x_{B} from the equation.

x_{B}=\frac{P- P_{T}^{o}}{ P_{B}^{o} - P_{T}^{o}}

x_{B}=\frac{35- 22}{75-22} = 0.24

So  

x_{T}=1-0.24 = 0.76

To get the mole fraction for the vapor we know that in the equilibrium:

P_{B}=y_{B}*P

y_{T}=1-y_{B}

So  

y_{B} =\frac{P_{B}}{P}=\frac{ x_{B}*P_{B}^{o}}{P}

y_{B}=\frac{0.24*75}{35}=0.51

y_{T}=1-0.51=0.49

Something that we can see in these compositions is that the liquid is richer in the less volatile compound (toluene) and the vapor in the more volatile compound (benzene). If we take away this vapor from the solution, the solution is going to reach a new state of equilibrium, where more vapor will be produced. This vapor will have a higher molar fraction of the more volatile compound. If we do this a lot of times, we can get a vapor that is almost pure in the more volatile compound. This is principle used in the fractional distillation.

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