the vapor pressure of ethanol at 20c is 44 mmHg and the vapor pressure of methanol at the same temperature is 94mmHg. A mixture of 30.0 g of methanol and 45.0 g of ethanol is prepared (and can be assumed to behave as an ideal solution). a.) calculate the vapor pressure of methanol and ethanol above this solution at 20c

Question

the vapor pressure of ethanol at 20c is 44 mmHg and the vapor pressure of methanol at the same temperature is 94mmHg. A mixture of 30.0 g of methanol and 45.0 g of ethanol is prepared (and can be assumed to behave as an ideal solution). a.) calculate the vapor pressure of methanol and ethanol above this solution at 20c

b.) calculate the mole fraction of ethanol and methanol in the vapor above this solution at 20c
c.) suggest a method for seperating the two components of this solution

Answer 1

moles MeOH = n= 30.0g/molar mass MeOH = ?

moles EtOH = n= 45.0g/molar mass EtOH = ?
XMeOH = nMeOH/total moles
XEtOH = nEtOH/total moles.
a.
PMeOH = XMeOH*P^o_MeOH
PEtOH = XEtOH*P^{o^{_EtOH

b.

total P above the soln is PMeOH + PEtOH.

XMeOH = PMeOH/total P

XEtOH = PEtOH/total P

c.Your call.}}

Answer 2

Thank u soooo much😭😭😭

Answer 3

To calculate the vapor pressure of methanol and ethanol above the solution, we can use Raoult's law, which states that the partial pressure of a component in an ideal solution is proportional to its mole fraction.

a) Calculating the vapor pressure of methanol above the solution:
Let's assume that the mole fraction of methanol in the solution is x.

The mole fraction of ethanol would be (1 - x) since the total mole fraction should add up to 1.
Given:
vapor pressure of methanol (Pmethanol) = 94 mmHg
vapor pressure of ethanol (Pethanol) = 44 mmHg

By Raoult's law, we have:
Pmethanol = x * 94 mmHg
Pethanol = (1 - x) * 44 mmHg

b) Calculating the mole fraction of ethanol and methanol in the vapor above the solution:
Using Raoult's law again, we can express the mole fraction of a component in the vapor phase as the ratio of the partial pressure of that component to the total vapor pressure.

Let Xethanol be the mole fraction of ethanol in the vapor phase and Xmethanol be the mole fraction of methanol in the vapor phase. The total vapor pressure is given by the sum of the partial pressures of both components:

Ptotal = Pethanol + Pmethanol

Using the equation above, we can find the mole fraction of each component in the vapor phase:

Xethanol = Pethanol / Ptotal
Xmethanol = Pmethanol / Ptotal

c) Suggesting a method for separating the two components of this solution:
To separate methanol and ethanol, we can use the process of fractional distillation. In fractional distillation, a mixture is heated to a boiling point at which one of the components vaporizes while the other remains in the liquid phase. The vapor is then cooled and condensed, resulting in separation of the components based on their different boiling points.

Since methanol has a lower boiling point (64.7°C) compared to ethanol (78.4°C), when the mixture is heated, methanol will vaporize first. The vapor is then collected and condensed, resulting in the separation of methanol from ethanol. This process enables the extraction of pure components from the solution.

Answer 4

a.) To calculate the vapor pressure of methanol and ethanol above the solution, we can use Raoult's law. Raoult's law states that the vapor pressure of the components in an ideal solution is proportional to the mole fraction of the component in the solution.

The mole fraction (X) of a component is the ratio of the number of moles of that component to the total number of moles in the solution. We can calculate the mole fraction of methanol and ethanol as follows:

Mole fraction of methanol (X_methanol) = moles of methanol / total moles
Mole fraction of ethanol (X_ethanol) = moles of ethanol / total moles

To find the moles of methanol and ethanol, we need to convert the given masses into moles using their molar masses. The molar mass of methanol (CH3OH) is 32.04 g/mol, and the molar mass of ethanol (C2H5OH) is 46.07 g/mol.

moles of methanol = 30.0 g / molar mass of methanol
moles of ethanol = 45.0 g / molar mass of ethanol

Now, we can calculate the mole fractions:

X_methanol = moles of methanol / (moles of methanol + moles of ethanol)
X_ethanol = moles of ethanol / (moles of methanol + moles of ethanol)

b.) The vapor pressure (P) of each component above the solution can be calculated using Raoult's law:

P_methanol = X_methanol * vapor pressure of methanol at 20°C
P_ethanol = X_ethanol * vapor pressure of ethanol at 20°C

Substituting the values given in the question:
P_methanol = X_methanol * 94 mmHg
P_ethanol = X_ethanol * 44 mmHg

c.) To separate the two components of this solution (methanol and ethanol), you can utilize the difference in boiling points between the two substances. Methanol has a lower boiling point (64.7°C) compared to ethanol (78.4°C). This means that if you heat the mixture, the methanol will evaporate first, leaving behind the ethanol in the liquid form. The vapor can then be cooled and condensed to obtain pure methanol, while the remaining liquid can be heated further to separate ethanol. This process is known as fractional distillation.