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An ideal solution was found to have a vapour pressure of 80 torr when the mole fraction of a non-volatile solute was 0.2. What would be the vapour pressure of the pure solvent at the same temperature?
Correct answer is '100'. Can you explain this answer?
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Solution:
To understand the relationship between the vapor pressure of an ideal solution and the mole fraction of a non-volatile solute, we can make use of Raoult's law.
Raoult's Law:
According to Raoult's law, the vapor pressure of an ideal solution is directly proportional to the mole fraction of the solvent present in the solution. Mathematically, it can be represented as:
P = P°solvent * Xsolvent
where P is the vapor pressure of the solution, P°solvent is the vapor pressure of the pure solvent, and Xsolvent is the mole fraction of the solvent.
Given:
P = 80 torr (vapor pressure of the ideal solution)
Xsolvent = 0.2 (mole fraction of the solvent)
We need to find the vapor pressure of the pure solvent (P°solvent).
Solution:
We can rearrange Raoult's law equation to solve for P°solvent:
P = P°solvent * Xsolvent
P°solvent = P / Xsolvent
Substituting the given values:
P°solvent = 80 torr / 0.2
P°solvent = 400 torr
Therefore, the vapor pressure of the pure solvent at the same temperature would be 400 torr.
However, it is important to note that the correct answer provided is '100 torr'. This suggests that there may be an error in the given information or an alternative explanation for the result. Without further clarification or additional information, it is difficult to determine the exact reasoning behind the correct answer.
To understand the relationship between the vapor pressure of an ideal solution and the mole fraction of a non-volatile solute, we can make use of Raoult's law.
Raoult's Law:
According to Raoult's law, the vapor pressure of an ideal solution is directly proportional to the mole fraction of the solvent present in the solution. Mathematically, it can be represented as:
P = P°solvent * Xsolvent
where P is the vapor pressure of the solution, P°solvent is the vapor pressure of the pure solvent, and Xsolvent is the mole fraction of the solvent.
Given:
P = 80 torr (vapor pressure of the ideal solution)
Xsolvent = 0.2 (mole fraction of the solvent)
We need to find the vapor pressure of the pure solvent (P°solvent).
Solution:
We can rearrange Raoult's law equation to solve for P°solvent:
P = P°solvent * Xsolvent
P°solvent = P / Xsolvent
Substituting the given values:
P°solvent = 80 torr / 0.2
P°solvent = 400 torr
Therefore, the vapor pressure of the pure solvent at the same temperature would be 400 torr.
However, it is important to note that the correct answer provided is '100 torr'. This suggests that there may be an error in the given information or an alternative explanation for the result. Without further clarification or additional information, it is difficult to determine the exact reasoning behind the correct answer.
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An ideal solution was found to have a vapour pressure of 80 torr when the mole fraction of a non-volatile solute was 0.2. What would be the vapour pressure of the pure solvent at the same temperature?Correct answer is '100'. Can you explain this answer?
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