If the vapor pressure of pure solvent is 120 mmHg, what will be the vapor pressure of a solution containing 0.2 moles of solute in 1 mole of solvent?
Practice Questions
1 question
Q1
If the vapor pressure of pure solvent is 120 mmHg, what will be the vapor pressure of a solution containing 0.2 moles of solute in 1 mole of solvent?
100 mmHg
110 mmHg
120 mmHg
80 mmHg
Using Raoult's Law, the vapor pressure of the solution is P_solution = P_solvent * X_solvent = 120 * (1/(1+0.2)) = 120 * (1/1.2) = 100 mmHg.
Questions & Step-by-step Solutions
1 item
Q
Q: If the vapor pressure of pure solvent is 120 mmHg, what will be the vapor pressure of a solution containing 0.2 moles of solute in 1 mole of solvent?
Solution: Using Raoult's Law, the vapor pressure of the solution is P_solution = P_solvent * X_solvent = 120 * (1/(1+0.2)) = 120 * (1/1.2) = 100 mmHg.
Steps: 7
Step 1: Identify the vapor pressure of the pure solvent, which is given as 120 mmHg.
Step 2: Determine the number of moles of solute and solvent. We have 0.2 moles of solute and 1 mole of solvent.
Step 3: Calculate the total number of moles in the solution. This is the sum of moles of solute and moles of solvent: 0.2 + 1 = 1.2 moles.
Step 4: Calculate the mole fraction of the solvent (X_solvent). This is the moles of solvent divided by the total moles: X_solvent = 1 / 1.2.
Step 5: Calculate the vapor pressure of the solution using Raoult's Law: P_solution = P_solvent * X_solvent.
Step 6: Substitute the values into the equation: P_solution = 120 mmHg * (1 / 1.2).
