The vapor pressure of the solvent decreased by 10 mm Hg, when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent, if the decrease in vapor pressure of the solvent is required to be 20 mm Hg. ?

ANS:D - 0.6

To solve this problem, we can use Raoult's Law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent in the solution. Mathematically, it can be represented as: 𝑃solution=𝑥solvent⋅𝑃solvent∗Psolution​=xsolvent​⋅Psolvent∗​ Where:

• 𝑃solutionPsolution​ is the vapor pressure of the solution,
• 𝑥solventxsolvent​ is the mole fraction of the solvent,
• 𝑃solvent∗Psolvent∗​ is the vapor pressure of the pure solvent.

Given that the vapor pressure of the solvent decreases by 10 mm Hg when the solute is added and the mole fraction of the solute is 0.2, the mole fraction of the solvent is initially 1−0.2=0.81−0.2=0.8. Now, we can use the concept that the change in vapor pressure is proportional to the mole fraction of the solvent. If the decrease in vapor pressure is required to be 20 mm Hg, then the decrease in mole fraction of the solvent should also be doubled. Initially, the decrease in vapor pressure is 10 mm Hg when the mole fraction of the solute is 0.2. To achieve a decrease of 20 mm Hg, the mole fraction of the solvent needs to be reduced by twice the amount, i.e., 0.2. So, the mole fraction of the solvent in the final solution would be 0.8−0.2=0.60.8−0.2=0.6. Therefore, the correct answer is: 0.6