A solution consists of 3.83 g benzene (C6H6) and 2.45 g toluene (C6H5CH3). The vapor pressure of pure benzene at 20°C is 75 mmHg and that of toluene is 22 mmHg. Assume Raoult's law holds. Calculate th

Molar mass of benzene ($\text{C}_6\text{H}_6$):

Molar mass of toluene ($\text{C}_6\text{H}_5\text{CH}_3$ or $\text{C}_7\text{H}_8$):

Moles of benzene ($n_{benzene}$):

Moles of toluene ($n_{toluene}$):

Total moles ($n_{total}$):

Mole fraction of benzene in liquid ($X_{benzene}$):

Mole fraction of toluene in liquid ($X_{toluene}$):

Given: $P_{benzene}^\circ = 75$ mmHg, $P_{toluene}^\circ = 22$ mmHg.
Partial pressure of benzene ($P_{benzene}$):

Partial pressure of toluene ($P_{toluene}$):

Total vapor pressure ($P_{total}$):

Mole fraction of benzene in vapor ($Y_{benzene}$):

Mole fraction of toluene in vapor ($Y_{toluene}$):

The mole fraction of benzene in the vapor is $\boxed{\text{0.863}}$ and the mole fraction of toluene in the vapor is $\boxed{\text{0.137}}$.

• Chloroform and methanol form an ideal solution. The solution boils at $22^\circ\text{C}$ and 0.255 atm. At $22^\circ\text{C}$, the vapor pressure of pure methanol is 0.192 atm and the vapor pressure of pure chloroform is 0.311 atm. What is the mole fraction of chloroform in solution?

Given:
Total vapor pressure, $P_{total} = 0.255$ atm
Vapor pressure of pure methanol, $P_{methanol}^\circ = 0.192$ atm
Vapor pressure of pure chloroform, $P_{chloroform}^\circ = 0.311$ atm
Unknown: Mole fraction of chloroform in the liquid solution, $X_{chloroform}$.

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

The mole fraction of chloroform in the solution is $\boxed{\text{0.529}}$.

• Heptane, $\text{C}_7\text{H}_{16}$ and octane $\text{C}_8\text{H}_{18}$, form ideal solution. At $40^\circ\text{C}$, the vapor pressure of pure heptane is 0.522 atm and the vapor pressure of pure octane is 0.238 atm. A solution is made of $5.32$g heptane and $8.80$g octane. Calculate the mole fraction of octane in the vapor at the above temperature.

Molar mass of heptane ($\text{C}_7\text{H}_{16}$):

Molar mass of octane ($\text{C}_8\text{H}_{18}$):

Moles of heptane ($n_{heptane}$):

Moles of octane ($n_{octane}$):

Total moles ($n_{total}$):

Mole fraction of heptane in liquid ($X_{heptane}$):

Mole fraction of octane in liquid ($X_{octane}$):

Given: $P_{heptane}^\circ = 0.522$ atm, $P_{octane}^\circ = 0.238$ atm.
Partial pressure of heptane ($P_{heptane}$):

Partial pressure of octane ($P_{octane}$):

Total vapor pressure ($P_{total}$):

Mole fraction of octane in vapor ($Y_{octane}$):

The mole fraction of octane in the vapor is $\boxed{\text{0.398}}$.