The vapour pressure of a solvent decreased by 10 mm Hg when a non-vola...
Explanation:
When a non-volatile solute is added to a solvent, it lowers the vapor pressure of the solvent. This is known as Raoult's law. According to Raoult's law, the vapor pressure of a solvent in a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.
Let's assume the initial vapor pressure of the pure solvent is P°. When the solute is added, the vapor pressure of the solvent decreases by 10 mm Hg. Therefore, the new vapor pressure of the solvent is P° - 10 mm Hg.
Using Raoult's law:
According to Raoult's law, the vapor pressure of the solvent in the solution can be represented as:
P(solvent) = X(solvent) * P°(solvent)
Where P(solvent) is the vapor pressure of the solvent in the solution, X(solvent) is the mole fraction of the solvent in the solution, and P°(solvent) is the vapor pressure of the pure solvent.
Given that the mole fraction of the solute in the solution is 0.2, we can represent the mole fraction of the solvent as:
X(solvent) = 1 - X(solute)
X(solvent) = 1 - 0.2
X(solvent) = 0.8
Now, we can use the equation P(solvent) = X(solvent) * P°(solvent) to find the vapor pressure of the solvent in the solution:
P(solvent) = 0.8 * P°(solvent)
Since the decrease in vapor pressure is 20 mm Hg, we can write the equation as:
P°(solvent) - 10 = 0.8 * P°(solvent)
Simplifying the equation:
0.2 * P°(solvent) = 10
P°(solvent) = 10 / 0.2
P°(solvent) = 50 mm Hg
Now, we can substitute this value back into the equation P(solvent) = 0.8 * P°(solvent) to find the mole fraction of the solvent:
P(solvent) = 0.8 * 50
P(solvent) = 40 mm Hg
Therefore, the mole fraction of the solvent in the solution when the decrease in vapor pressure is 20 mm Hg is 0.6 (option B).