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Rom4ik [11]
3 years ago
5

What amount of a nonelectrolytic solute should be dissolved in 28.9 moles of benzene (C₆H₆) at 25°C to change the vapor pressure

of benzene by 17.9%? The vapor pressure of pure benzene at 25°C is 94.2 torr.
Chemistry
1 answer:
RideAnS [48]3 years ago
8 0

Answer:

The amount of a nonelectrolytic solute that has to be dissolved in 28.9 moles to change the vapor pressure of benzene by 17.9 is 6.31 moles.

Explanation:

Let's see how to solve this, step by step applying Raoult's Law (Colligative property).

Pressure solution = χsolvent . Pressure solvent

You have to consider that Pressure in solution is, Pressure Pure - Pressure in Solution (Solute+Solvent). The statement says, that pressure has fallen 17.9%, so let's find out that

100% ....... 94.2 Torr

17.9 % ...... x

16.86 Torr (after the Rule of 3) - This is the value of decline.

So, 94.2Torr - 16.86 Torr = 77.34 Torr

Now the formula

77.34 Torr = 94.2 Torr . X (where x is molar fraction)

x = moles of benzene / moles of benzene + moles of nonelectrolytic solute

77.34 Torr / 94.2 Torr = X

0.821 = moles of benzene / moles of benzene + moles of nonelectrolytic solute

0.821 = 28.9 moles / 28.9 moles + nonelectrolytic

0.821 (28.9 moles + nonelectrolytic) = 28.9 moles

23.72 moles + 0.821 nonelectrolytic = 28.9 moles

0.821 nonelectrolytic = 28.9 moles - 23.72 moles

0.821 nonelectrolytic = 5.18 moles

nonelectrolytic moles = 5.18 / 0.821

nonelectrolytic moles = 6.31

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sp2606 [1]

The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

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The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

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<u>Explanation:</u>

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                  1.8                4.7

<u>At eqllm:</u>           1.8-x             4.7-x

Evaluating the value of 'x'

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The expression of K_p for above equation follows:

K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}

p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm

Putting values in above expression, we get:

K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136

When more ethylene is added, the equilibrium gets re-established.

Partial pressure of ethylene added = 1.2 atm

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                2.36             4.06               0.64

<u>At eqllm:</u>           2.36-x        4.06-x             0.64+x

Putting value in the equilibrium constant expression, we get:

0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41

Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.

So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm

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