This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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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 = 1 For X_B = 0.25: X_M = 1 - 0.25 = 0.75 For X_B = 0.5: X_M = 1 - 0.5 = 0.5 For X_B = 0.75: X_M = 1 - 0.75 = 0.25 For X_B = 1: X_M = 1 - 1 = 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 · P_B^0. For X_B = 0: P_B = 0 × 95.1 mmHg = 0 mmHg For X_B = 0.25: P_B = 0.25 × 95.1 mmHg = 23.78 mmHg For X_B = 0.5: P_B = 0.5 × 95.1 mmHg = 47.55 mmHg For X_B = 0.75: P_B = 0.75 × 95.1 mmHg = 71.33 mmHg For X_B = 1: P_B = 1 × 95.1 mmHg = 95.1 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 · P_M^0. For X_M = 1 (when X_B = 0): P_M = 1 × 28.4 mmHg = 28.4 mmHg For X_M = 0.75 (when X_B = 0.25): P_M = 0.75 × 28.4 mmHg = 21.3 mmHg For X_M = 0.5 (when X_B = 0.5): P_M = 0.5 × 28.4 mmHg = 14.2 mmHg For X_M = 0.25 (when X_B = 0.75): P_M = 0.25 × 28.4 mmHg = 7.1 mmHg For X_M = 0 (when X_B = 1): P_M = 0 × 28.4 mmHg = 0 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 mmHg + 28.4 mmHg = 28.4 mmHg For X_B = 0.25: P_total = 23.78 mmHg + 21.3 mmHg = 45.08 mmHg For X_B = 0.5: P_total = 47.55 mmHg + 14.2 mmHg = 61.75 mmHg For X_B = 0.75: P_total = 71.33 mmHg + 7.1 mmHg = 78.43 mmHg For X_B = 1: P_total = 95.1 mmHg + 0 mmHg = 95.1 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