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The vapour pressure of a solvent decreased by 10 mm of Hg when a non volatile solute was added to the solvent. The mole fraction of solute in solution is 0.2, what would be the mole fraction of solvent if decrease in vapour pressure is 20 mm of Hg?
- a)0.8
- b)0.6
- c)0.4
- d)0.2
Correct answer is option 'C'. Can you explain this answer?
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Solution:
Given:
∆P = 10 mm of Hg (decrease in vapour pressure)
x2 = 0.2 (mole fraction of solute)
To find:
x1 (mole fraction of solvent) when ∆P = 20 mm of Hg
Formula Used:
Relative lowering of the vapour pressure (P°-P)/P° = x2
Where,
P° = Vapour pressure of pure solvent
P = Vapour pressure of solution
x2 = Mole fraction of solute
Derivation:
Let us assume that P° and P are the vapour pressures of the pure solvent and the solution respectively.
According to Raoult's law, the vapour pressure of a solution is directly proportional to the mole fraction of the solvent.
P = P° x x1
Where x1 is the mole fraction of the solvent.
If a non-volatile solute is added to the solvent, it decreases the vapour pressure of the solvent. The decrease in vapour pressure (∆P) is directly proportional to the mole fraction of the solute.
∆P/P° = x2
∆P = P° x x2
Putting the value of ∆P in the above equation, we get:
P = P° - P° x x2
P = P° (1 - x2)
Now, we are given that ∆P = 10 mm of Hg when x2 = 0.2
∆P/P° = 0.2
∆P = 0.2 P°
Putting the value of ∆P in the above equation, we get:
P° (1 - x2) = 0.8 P°
1 - x2 = 0.8
x2 = 0.2
Now, we need to find x1 when ∆P = 20 mm of Hg.
∆P/P° = x2 = 0.2
∆P = 0.2 P°
We know that P = P° (1 - x2)
Putting the value of ∆P in the above equation, we get:
P° (1 - x2) = P° - ∆P
P° (1 - x2) = P° - 0.2 P°
P° (1 - x2) = 0.8 P°
1 - x2 = 0.8
x1 = 0.2
Therefore, the mole fraction of the solvent when the decrease in vapour pressure is 20 mm of Hg is 0.4 (1 - 0.2).
Hence, option (C) is the correct answer.
Given:
∆P = 10 mm of Hg (decrease in vapour pressure)
x2 = 0.2 (mole fraction of solute)
To find:
x1 (mole fraction of solvent) when ∆P = 20 mm of Hg
Formula Used:
Relative lowering of the vapour pressure (P°-P)/P° = x2
Where,
P° = Vapour pressure of pure solvent
P = Vapour pressure of solution
x2 = Mole fraction of solute
Derivation:
Let us assume that P° and P are the vapour pressures of the pure solvent and the solution respectively.
According to Raoult's law, the vapour pressure of a solution is directly proportional to the mole fraction of the solvent.
P = P° x x1
Where x1 is the mole fraction of the solvent.
If a non-volatile solute is added to the solvent, it decreases the vapour pressure of the solvent. The decrease in vapour pressure (∆P) is directly proportional to the mole fraction of the solute.
∆P/P° = x2
∆P = P° x x2
Putting the value of ∆P in the above equation, we get:
P = P° - P° x x2
P = P° (1 - x2)
Now, we are given that ∆P = 10 mm of Hg when x2 = 0.2
∆P/P° = 0.2
∆P = 0.2 P°
Putting the value of ∆P in the above equation, we get:
P° (1 - x2) = 0.8 P°
1 - x2 = 0.8
x2 = 0.2
Now, we need to find x1 when ∆P = 20 mm of Hg.
∆P/P° = x2 = 0.2
∆P = 0.2 P°
We know that P = P° (1 - x2)
Putting the value of ∆P in the above equation, we get:
P° (1 - x2) = P° - ∆P
P° (1 - x2) = P° - 0.2 P°
P° (1 - x2) = 0.8 P°
1 - x2 = 0.8
x1 = 0.2
Therefore, the mole fraction of the solvent when the decrease in vapour pressure is 20 mm of Hg is 0.4 (1 - 0.2).
Hence, option (C) is the correct answer.
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The vapour pressure of a solvent decreased by 10 mm of Hg when a non volatile solute was added to the solvent. The mole fraction of solute in solution is 0.2, what would be the mole fraction of solvent if decrease in vapour pressure is 20 mm of Hg?a)0.8b)0.6c)0.4d)0.2Correct answer is option 'C'. Can you explain this answer?
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