1. **Problem Statement:** Calculate the vapor pressure of pure propanol B at 298 K given a mixture of 0.2 mole methanol A and 0.5 mole propanol B with total vapor pressure 40 mmHg, pressure of pure A is 20 mmHg, and the mixture obeys Raoult's law. 2. **Raoult's Law:** Raoult's law states that the partial vapor pressure of each component in an ideal solution is proportional to its mole fraction and the vapor pressure of the pure component: $$P_i = x_i P_i^*$$ where $P_i$ is the partial vapor pressure of component $i$, $x_i$ is the mole fraction of component $i$, and $P_i^*$ is the vapor pressure of pure component $i$. 3. **Calculate mole fractions:** Total moles $= 0.2 + 0.5 = 0.7$ Mole fraction of A: $$x_A = \frac{0.2}{0.7} = \frac{2}{7} \approx 0.2857$$ Mole fraction of B: $$x_B = \frac{0.5}{0.7} = \frac{5}{7} \approx 0.7143$$ 4. **Apply Raoult's law for total pressure:** Total vapor pressure $P_{total} = P_A + P_B = x_A P_A^* + x_B P_B^*$ Given: $$P_{total} = 40 \text{ mmHg}$$ $$P_A^* = 20 \text{ mmHg}$$ Substitute values: $$40 = 0.2857 \times 20 + 0.7143 \times P_B^*$$ Calculate partial pressure of A: $$0.2857 \times 20 = 5.714$$ So: $$40 = 5.714 + 0.7143 P_B^*$$ 5. **Solve for $P_B^*$:** $$0.7143 P_B^* = 40 - 5.714 = 34.286$$ $$P_B^* = \frac{34.286}{0.7143} \approx 48.0 \text{ mmHg}$$ **Final answer:** The vapor pressure of pure propanol B at 298 K is approximately **48.0 mmHg**.