In a solution of two volatile 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 and that of pure B is 50 mmHg?
Practice Questions
1 question
Q1
In a solution of two volatile 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 and that of pure B is 50 mmHg?
80 mmHg
90 mmHg
70 mmHg
60 mmHg
Using Raoult's Law, the vapor pressure of the solution is P_total = (0.6 * 100) + (0.4 * 50) = 60 + 20 = 80 mmHg.
Questions & Step-by-step Solutions
1 item
Q
Q: In a solution of two volatile 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 and that of pure B is 50 mmHg?
Solution: Using Raoult's Law, the vapor pressure of the solution is P_total = (0.6 * 100) + (0.4 * 50) = 60 + 20 = 80 mmHg.
Steps: 7
Step 1: Identify the mole fraction of component A, which is given as 0.6.
Step 2: Calculate the mole fraction of component B. Since the total mole fraction must equal 1, the mole fraction of B is 1 - 0.6 = 0.4.
Step 3: Identify the vapor pressure of pure component A, which is given as 100 mmHg.
Step 4: Identify the vapor pressure of pure component B, which is given as 50 mmHg.
Step 5: Use Raoult's Law to calculate the partial vapor pressure of A in the solution: Partial pressure of A = mole fraction of A * vapor pressure of pure A = 0.6 * 100 mmHg = 60 mmHg.
Step 6: Use Raoult's Law to calculate the partial vapor pressure of B in the solution: Partial pressure of B = mole fraction of B * vapor pressure of pure B = 0.4 * 50 mmHg = 20 mmHg.
Step 7: Add the partial pressures of A and B to find the total vapor pressure of the solution: P_total = Partial pressure of A + Partial pressure of B = 60 mmHg + 20 mmHg = 80 mmHg.
