Liquids A and B form an ideal solution. The vapor pressure of pure A is 0.700 atm at the normal boiling point of a solution prepared from 0.250 mole of B and 0.650 mole of A. What is the vapor pressur

Given:
Vapor pressure of pure A, $P_A^\circ = 0.700$ atm
Moles of B, $n_B = 0.250$ mol
Moles of A, $n_A = 0.650$ mol
Since the solution is at its normal boiling point, the total vapor pressure, $P_{total} = 1.000$ atm.
Unknown: Vapor pressure of pure B, $P_B^\circ$.

Total moles, $n_{total} = n_A + n_B$

Mole fraction of A, $X_A = \frac{n_A}{n_{total}}$

Mole fraction of B, $X_B = \frac{n_B}{n_{total}}$

Raoult's Law states that $P_{total} = X_A P_A^\circ + X_B P_B^\circ$.
Substitute the known values into the equation:

The vapor pressure of pure B at this temperature is $\boxed{\text{1.78 atm}}$.

b) Acetone and methanol form an ideal solution at $25^\circ\text{C}$. The vapor pressure of pure acetone and pure methanol are 0.342 atm and 0.186 atm respectively. Calculate the mole fraction of methanol in a solution that boils at $25^\circ\text{C}$ and 0.248 atm.

Given:
Vapor pressure of pure acetone, $P_{acetone}^\circ = 0.342$ atm
Vapor pressure of pure methanol, $P_{methanol}^\circ = 0.186$ atm
Total vapor pressure of the solution, $P_{total} = 0.248$ atm
Unknown: Mole fraction of methanol, $X_{methanol}$.

Let $X_{methanol}$ be the mole fraction of methanol.
Then the mole fraction of acetone, $X_{acetone} = 1 - X_{methanol}$.
Raoult's Law states: $P_{total} = X_{acetone} P_{acetone}^\circ + X_{methanol} P_{methanol}^\circ$.
Substitute the known values into the equation:

Rounding to three significant figures:

The mole fraction of methanol in the solution is $\boxed{\text{0.603}}$.