At 20◦C the vapor pressure of pure methanol

(CH3OH) is 89.0 torr and of pure water (H2O)
is 17.5 torr. A solution of methanol and water
has a total vapor pressure of 35.5 torr at
20◦C. Calculate the mole fraction of methanol
in this solution. Assume the solution obeys
Raoult’s Law

I haven't a clue what to do for this problem. Wasn't in class for example problems so I don't know what goes with what to determine what.

We can't draw diagrams on this forum but it is easy to see if I could draw a graph.

Basically this is what you do.
XH2O = Pomethanol-Ptotal/(Pomethanol-PoH2O)
That gives you (89.0-35.5)/(89.0 - 17.5) = 0.748 for H2O

I can never remember the formula; I draw it out with a graph and do it by proportion.
Then Xmethanol = 1-XH2O or you can do it for methanol first as
Xme = (Ptotal-PoH2O)/(Pomethanol - PoH2O) . That should give you the same answer as 1-0.748 = ?

No worries! I can help you solve this problem step by step.

To find the mole fraction of methanol in the solution, we first need to understand Raoult's Law. Raoult's Law states that the vapor pressure of a solution is directly proportional to the mole fraction of each component in the solution.

The formula for Raoult's Law is:

P(total) = X(methanol) * P(methanol) + X(water) * P(water)

Where P(total) is the total vapor pressure of the solution, X(methanol) is the mole fraction of methanol, P(methanol) is the vapor pressure of pure methanol, X(water) is the mole fraction of water, and P(water) is the vapor pressure of pure water.

In this problem, we are given P(total) = 35.5 torr, P(methanol) = 89.0 torr, and P(water) = 17.5 torr. We still need to find X(methanol).

Let's start by rearranging the equation to solve for X(methanol):

X(methanol) = (P(total) - X(water) * P(water)) / P(methanol)

Now, we need to find X(water). The mole fraction of water can be calculated using the equation:

X(water) = 1 - X(methanol)

Substituting this expression into the previous equation, we get:

X(methanol) = (P(total) - (1 - X(methanol)) * P(water)) / P(methanol)

Simplifying further:

X(methanol) = (P(total) - P(water) + X(methanol) * P(water)) / P(methanol)

Now, let's solve this equation for X(methanol):

X(methanol) * P(methanol) - X(methanol) * P(water) = P(total) - P(water)

Factoring out X(methanol) on the left-hand side:

X(methanol) * (P(methanol) - P(water)) = P(total) - P(water)

Finally, we can solve for X(methanol):

X(methanol) = (P(total) - P(water)) / (P(methanol) - P(water))

Now you can plug in the values you have and calculate the mole fraction of methanol in the solution.

No worries, I can help you step-by-step with this problem! To calculate the mole fraction of methanol in the solution, we need to use Raoult's Law. Raoult's Law states that for an ideal solution, the vapor pressure of a component is equal to the mole fraction of that component multiplied by its vapor pressure at that temperature. Let's break it down:

Step 1: Write down the given information:
- Vapor pressure of pure methanol (CH3OH) = 89.0 torr
- Vapor pressure of pure water (H2O) = 17.5 torr
- Total vapor pressure of the solution = 35.5 torr

Step 2: Set up the equation using Raoult's Law:
Total vapor pressure = mole fraction of methanol x vapor pressure of methanol + mole fraction of water x vapor pressure of water

Step 3: Assign variables:
Let's use x for the mole fraction of methanol and (1-x) for the mole fraction of water. The mole fractions of both components add up to 1, so the mole fraction of water is (1 - x).

Step 4: Substitute the given information and variables into Raoult's Law equation:
35.5 torr = x * 89.0 torr + (1 - x) * 17.5 torr

Step 5: Simplify and solve for x:
35.5 torr = 89.0x + 17.5 - 17.5x
35.5 torr - 17.5 torr = 71.5x
18 torr = 71.5x
x = 18 torr / 71.5 torr

Step 6: Calculate x:
Divide both sides by 71.5:
x ≈ 0.252 (rounded to three decimal places)

Step 7: Interpret the result:
The mole fraction of methanol in the solution is approximately 0.252.

I hope this step-by-step explanation helps! Let me know if you have any further questions.