148.0 grams of ether are mixed with 184.0 grams of ethanol at 80 degrees Celsius. The vapor pressure of ether at 80 degrees Celsius is 4.0 and for ethanol at 80 degrees Celsius is 1.1. Determine the vapor pressure of the solution

Solution

To solve this question,we first need to think about the Raoult’s Law which states that the partial vapor pressure of a component in a mixture is equal to the vapor pressure of the pure component at that temperature multiplied by its mole fraction in the mixture.

This means that we need to find the partial pressure for each component of the mixture the add them .

P ether = X ether x Po ether

P ethanol = X ethanol x Po ethanol

Total vapor pressure = P ether + P ethanol

Where P is the partial vapor pressure of the component;X is the mole fraction of the components and Po is the value of vapor pressures of the components when on their own.

Now calculating the moles

moles of ether = 148/74.12 = 1.97 moles

moles of ethanol= 184/46.07 = 3.99 moles

total moles = 1.97 + 3.99 = 5.96 moles

X ether= moles of ether/total moles

X ether = 1.97 / 5.96 = 0.331

X ethanol = 3.99/5.96 = 0.669

Also given;

Poether = 4.0

Poethanol = 1.1

P ether = 0.331 x 4.0 = 1.324

P ethanol = 0.669 x 1.1 = 0.7359

Total vapor pressure = 1.324 + 0.7359 = 2.0599 = 2.06

Thus the vapor pressure of the solution is 2.06

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