A solution is made by mixing benzene and methylbenzene (toluene). If the mole fractions of benzene are 0, 0.25, 0.5, 0.75, and 1, calculate the mole fraction of methylbenzene and the partial vapor pre

To solve this problem, we will use Raoult's Law for ideal solutions. We need the pure vapor pressures of Benzene and Methylbenzene (Toluene).

Assumptions:

• The solution is ideal.
• The temperature is constant, typically 25°C.
• Pure vapor pressure of Benzene ($P_B^0$) = 95.1 mmHg
• Pure vapor pressure of Methylbenzene (Toluene) ($P_M^0$) = 28.4 mmHg

The mole fractions of Benzene ($X_B$) are given as: 0, 0.25, 0.5, 0.75, and 1.

---

a. Find the respective mole fraction of methylbenzene at this concentration.
The sum of mole fractions in a binary solution is 1: $X_B + X_M = 1$.
So, $X_M = 1 - X_B$.

• For $X_B = 0$: $X_M = 1 - 0 = \boxed{1}$
• For $X_B = 0.25$: $X_M = 1 - 0.25 = \boxed{0.75}$
• For $X_B = 0.5$: $X_M = 1 - 0.5 = \boxed{0.5}$
• For $X_B = 0.75$: $X_M = 1 - 0.75 = \boxed{0.25}$
• For $X_B = 1$: $X_M = 1 - 1 = \boxed{0}$

---

bi. Calculate the respective vapour pressure contributions of benzene at these concentration.
According to Raoult's Law, the partial vapor pressure of Benzene ($P_B$) is given by $P_B = X_B \cdot P_B^0$.

• For $X_B = 0$: $P_B = 0 \times 95.1 \text{ mmHg} = \boxed{0 \text{ mmHg}}$
• For $X_B = 0.25$: $P_B = 0.25 \times 95.1 \text{ mmHg} = \boxed{23.78 \text{ mmHg}}$
• For $X_B = 0.5$: $P_B = 0.5 \times 95.1 \text{ mmHg} = \boxed{47.55 \text{ mmHg}}$
• For $X_B = 0.75$: $P_B = 0.75 \times 95.1 \text{ mmHg} = \boxed{71.33 \text{ mmHg}}$
• For $X_B = 1$: $P_B = 1 \times 95.1 \text{ mmHg} = \boxed{95.1 \text{ mmHg}}$

---

bii. Calculate the respective vapour pressure contributions for methylbenzene at these concentration.
According to Raoult's Law, the partial vapor pressure of Methylbenzene ($P_M$) is given by $P_M = X_M \cdot P_M^0$.

• For $X_M = 1$ (when $X_B = 0$): $P_M = 1 \times 28.4 \text{ mmHg} = \boxed{28.4 \text{ mmHg}}$
• For $X_M = 0.75$ (when $X_B = 0.25$): $P_M = 0.75 \times 28.4 \text{ mmHg} = \boxed{21.3 \text{ mmHg}}$
• For $X_M = 0.5$ (when $X_B = 0.5$): $P_M = 0.5 \times 28.4 \text{ mmHg} = \boxed{14.2 \text{ mmHg}}$
• For $X_M = 0.25$ (when $X_B = 0.75$): $P_M = 0.25 \times 28.4 \text{ mmHg} = \boxed{7.1 \text{ mmHg}}$
• For $X_M = 0$ (when $X_B = 1$): $P_M = 0 \times 28.4 \text{ mmHg} = \boxed{0 \text{ mmHg}}$

---

c. Calculate the total Vapour pressures at the Various concentration.
According to Dalton's Law of Partial Pressures, the total vapor pressure ($P_{total}$) is the sum of the partial vapor pressures: $P_{total} = P_B + P_M$.

• For $X_B = 0$: $P_{total} = 0 \text{ mmHg} + 28.4 \text{ mmHg} = \boxed{28.4 \text{ mmHg}}$
• For $X_B = 0.25$: $P_{total} = 23.78 \text{ mmHg} + 21.3 \text{ mmHg} = \boxed{45.08 \text{ mmHg}}$
• For $X_B = 0.5$: $P_{total} = 47.55 \text{ mmHg} + 14.2 \text{ mmHg} = \boxed{61.75 \text{ mmHg}}$
• For $X_B = 0.75$: $P_{total} = 71.33 \text{ mmHg} + 7.1 \text{ mmHg} = \boxed{78.43 \text{ mmHg}}$
• For $X_B = 1$: $P_{total} = 95.1 \text{ mmHg} + 0 \text{ mmHg} = \boxed{95.1 \text{ mmHg}}$

---

Summary Table for Plotting (d):

$X_B$	$X_M$	$P_B$ (mmHg)	$P_M$ (mmHg)	$P_{total}$ (mmHg)
0	1	0	28.4	28.4
0.25	0.75	23.78	21.3	45.08
0.5	0.5	47.55	14.2