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Two liquid X and Y form an ideal solutio...

Two liquid X and Y form an ideal solution. At 300K vapour pressure of the solution containing 1 mol of X and 3 mol of Y 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mmHg) of X and Y in their pure states will be , respectively :

A

300 and 400

B

400 and 600

C

500 and 600

D

200 and 300

Text Solution

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To solve the problem step by step, we will follow the information provided in the question and the video transcript. ### Step 1: Identify the initial conditions We have two liquids, X and Y, forming an ideal solution. Initially, we have: - 1 mol of X - 3 mol of Y The total vapor pressure of this solution at 300 K is given as 550 mm Hg. ### Step 2: Calculate mole fractions The mole fraction of X (χX) and Y (χY) can be calculated as follows: - Total moles = 1 (X) + 3 (Y) = 4 - Mole fraction of X (χX) = moles of X / total moles = 1/4 - Mole fraction of Y (χY) = moles of Y / total moles = 3/4 ### Step 3: Write the equation for total vapor pressure Using Raoult's Law, the total vapor pressure (P) of the solution can be expressed as: \[ P = P_X \cdot χ_X + P_Y \cdot χ_Y \] Where: - \( P_X \) = vapor pressure of X in pure state - \( P_Y \) = vapor pressure of Y in pure state Substituting the known values: \[ 550 = P_X \cdot \frac{1}{4} + P_Y \cdot \frac{3}{4} \] ### Step 4: Rearranging the equation Multiplying through by 4 to eliminate the fractions: \[ 2200 = P_X + 3P_Y \] This is our **Equation 1**. ### Step 5: Analyze the second condition When 1 mol of Y is added, the new moles become: - 1 mol of X - 4 mol of Y Total moles = 1 + 4 = 5. The new vapor pressure of the solution is 560 mm Hg (an increase of 10 mm Hg). ### Step 6: Calculate new mole fractions - Mole fraction of X (χX) = 1/5 - Mole fraction of Y (χY) = 4/5 ### Step 7: Write the equation for the new total vapor pressure Using Raoult's Law again: \[ 560 = P_X \cdot \frac{1}{5} + P_Y \cdot \frac{4}{5} \] ### Step 8: Rearranging the new equation Multiplying through by 5: \[ 2800 = P_X + 4P_Y \] This is our **Equation 2**. ### Step 9: Solve the system of equations Now we have two equations: 1. \( P_X + 3P_Y = 2200 \) (Equation 1) 2. \( P_X + 4P_Y = 2800 \) (Equation 2) Subtract Equation 1 from Equation 2: \[ (P_X + 4P_Y) - (P_X + 3P_Y) = 2800 - 2200 \] This simplifies to: \[ P_Y = 600 \] ### Step 10: Substitute \( P_Y \) back to find \( P_X \) Substituting \( P_Y \) into Equation 1: \[ P_X + 3(600) = 2200 \] \[ P_X + 1800 = 2200 \] \[ P_X = 2200 - 1800 = 400 \] ### Final Answer The vapor pressures of X and Y in their pure states are: - \( P_X = 400 \) mm Hg - \( P_Y = 600 \) mm Hg
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