If the vapor pressure of a pure solvent is 80 mmHg and a non-volatile solute is added, resulting in a vapor pressure of 60 mmHg, what is the mole fraction of the solvent in the solution?
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If the vapor pressure of a pure solvent is 80 mmHg and a non-volatile solute is added, resulting in a vapor pressure of 60 mmHg, what is the mole fraction of the solvent in the solution?
Q: If the vapor pressure of a pure solvent is 80 mmHg and a non-volatile solute is added, resulting in a vapor pressure of 60 mmHg, what is the mole fraction of the solvent in the solution?
Step 1: Understand that the vapor pressure of a pure solvent is given as 80 mmHg.
Step 2: Recognize that when a non-volatile solute is added, the vapor pressure decreases to 60 mmHg.
Step 3: Recall Raoult's Law, which states that the vapor pressure of the solution (P_solution) is equal to the mole fraction of the solvent (X_solvent) multiplied by the vapor pressure of the pure solvent (P_pure_solvent).
Step 4: Write the equation from Raoult's Law: P_solution = X_solvent * P_pure_solvent.
Step 5: Substitute the known values into the equation: 60 mmHg = X_solvent * 80 mmHg.
Step 6: To find X_solvent, rearrange the equation: X_solvent = 60 mmHg / 80 mmHg.
