For a solution containing 2 components A and B, if the mole fraction of A is 0.6, what is the vapor pressure of the solution if the vapor pressure of pure A is 100 mmHg?
Practice Questions
1 question
Q1
For a solution containing 2 components A and B, if the mole fraction of A is 0.6, what is the vapor pressure of the solution if the vapor pressure of pure A is 100 mmHg?
60 mmHg
100 mmHg
40 mmHg
80 mmHg
According to Raoult's Law, the vapor pressure of the solution is P_A = X_A * P_A^0 = 0.6 * 100 mmHg = 60 mmHg.
Questions & Step-by-step Solutions
1 item
Q
Q: For a solution containing 2 components A and B, if the mole fraction of A is 0.6, what is the vapor pressure of the solution if the vapor pressure of pure A is 100 mmHg?
Solution: According to Raoult's Law, the vapor pressure of the solution is P_A = X_A * P_A^0 = 0.6 * 100 mmHg = 60 mmHg.
Steps: 6
Step 1: Understand that the mole fraction of A (X_A) is given as 0.6.
Step 2: Know that the vapor pressure of pure A (P_A^0) is given as 100 mmHg.
Step 3: Recall Raoult's Law, which states that the vapor pressure of the solution (P_A) is equal to the mole fraction of A (X_A) multiplied by the vapor pressure of pure A (P_A^0).
Step 4: Substitute the values into the formula: P_A = X_A * P_A^0.
Step 5: Calculate P_A by multiplying 0.6 (X_A) by 100 mmHg (P_A^0).
Step 6: The result is P_A = 0.6 * 100 mmHg = 60 mmHg.
