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Vapour pressure of solution obtained by ...

Vapour pressure of solution obtained by mixing 1 mole of n hexane and 3 mole of n-heptane is 550 mm Hg . On mixing 1 mole n-heptane, vapour pressure of solution increases by 10mm Hg. Find the vapour pressure of pure n-heptane

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To find the vapor pressure of pure n-heptane, we will follow these steps: ### Step 1: Understand the given data We have: - 1 mole of n-hexane (denote as A) - 3 moles of n-heptane (denote as B) - The vapor pressure of the solution with 1 mole of n-hexane and 3 moles of n-heptane is 550 mm Hg. - When we add 1 mole of n-heptane to this solution, the vapor pressure increases by 10 mm Hg, making it 560 mm Hg. ### Step 2: Calculate the mole fractions For the initial solution (1 mole of n-hexane and 3 moles of n-heptane): - Total moles = 1 + 3 = 4 - Mole fraction of n-hexane (XA) = nA / (nA + nB) = 1 / 4 = 0.25 - Mole fraction of n-heptane (XB) = nB / (nA + nB) = 3 / 4 = 0.75 ### Step 3: Write the equation for the vapor pressure of the initial solution Using Raoult's law: \[ P_{solution} = X_A P_A^0 + X_B P_B^0 \] Substituting the values we have: \[ 550 = (0.25) P_A^0 + (0.75) P_B^0 \] This can be rearranged to: \[ 550 = 0.25 P_A^0 + 0.75 P_B^0 \] Multiply through by 4 to eliminate the fraction: \[ 2200 = P_A^0 + 3 P_B^0 \] (Equation 1) ### Step 4: Analyze the new solution after adding 1 mole of n-heptane Now, we have: - 1 mole of n-hexane - 4 moles of n-heptane (3 original + 1 added) - Total moles = 1 + 4 = 5 - New mole fraction of n-hexane (XA) = 1 / 5 = 0.2 - New mole fraction of n-heptane (XB) = 4 / 5 = 0.8 ### Step 5: Write the equation for the vapor pressure of the new solution The new vapor pressure is 560 mm Hg: \[ 560 = (0.2) P_A^0 + (0.8) P_B^0 \] Multiply through by 5: \[ 2800 = P_A^0 + 4 P_B^0 \] (Equation 2) ### Step 6: Solve the equations simultaneously Now we have two equations: 1. \( 2200 = P_A^0 + 3 P_B^0 \) 2. \( 2800 = P_A^0 + 4 P_B^0 \) Subtract Equation 1 from Equation 2: \[ (2800 - 2200) = (P_A^0 + 4 P_B^0) - (P_A^0 + 3 P_B^0) \] This simplifies to: \[ 600 = P_B^0 \] ### Step 7: Conclusion The vapor pressure of pure n-heptane (P_B^0) is 600 mm Hg.
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