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The vapour pressure of pure benzene at 353 K is 10.0268 x 104 Nm-2 and that of a solution containing non-volatile solute in benzene is 9.89600 x 104 Nm-2. (d of benzene = 8.149 x 102 g dm-3). Thus, osmotic pressure of the solution is
- a)3.94 x 105 Nm-2
- b)4.00 x 105 Nm-2
- c)3.94 x 106 Nm-2
- d)4.00 x 106 Nm-2
Correct answer is option 'A'. Can you explain this answer?
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The vapour pressure of pure benzene at 353 K is 10.0268 x 104 Nm-2 and...
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The vapour pressure of pure benzene at 353 K is 10.0268 x 104 Nm-2 and...
Given data:
Vapour pressure of pure benzene at 353 K, P₁ = 10.0268 × 10⁴ Nm⁻²
Vapour pressure of solution, P₂ = 9.8960 × 10⁴ Nm⁻²
Density of benzene, d = 8.149 × 10² g dm⁻³
To find: Osmotic pressure of the solution
Formula used:
For a dilute solution, the osmotic pressure (π) is given by the equation:
π = (n/V)×RT
where n is the number of moles of the solute, V is the volume of the solution, R is the gas constant and T is the temperature.
The number of moles of the solute can be calculated as:
n = (m/M)
where m is the mass of the solute and M is its molar mass.
Solution:
Molar mass of benzene, C₆H₆ = 6 × 12.01 + 6 × 1.008 = 78.11 g mol⁻¹
Let us assume that we have 1000 g of the solution.
Then, the mass of benzene in the solution = (1000 - m) g
From Raoult's law, we know that:
P₂ = X₂ × P₁
where X₂ is the mole fraction of benzene in the solution.
Mole fraction of benzene in the solution can be calculated as:
X₂ = moles of benzene / total moles
Let the moles of benzene be n₂.
Then, the total moles can be calculated as:
n₁ + n₂ = (mass of benzene / Molar mass of benzene) + (mass of solute / Molar mass of solute)
Since the solute is non-volatile, its vapour pressure is negligible and hence its mole fraction can be assumed to be equal to 1.
Therefore, the above equation can be simplified as:
n₂ = (mass of benzene / Molar mass of benzene)
Thus, we have:
X₂ = n₂ / (n₁ + n₂)
X₂ = (mass of benzene / Molar mass of benzene) / [(mass of benzene / Molar mass of benzene) + (mass of solute / Molar mass of solute)]
Substituting the given values, we get:
9.8960 × 10⁴ = [(1000 - m) / 78.11] / [m / (M × 78.11)] × 10.0268 × 10⁴
Solving for m, we get:
m = 39.24 g
Mass of solute, w = (1000 - m) g = 960.76 g
Number of moles of the solute, n = (w / M) = (960.76 / M) mol
Volume of the solution, V = (w / d) = (960.76 / 8.149) dm³ = 117.8 dm³
Substituting the given values in the equation for osmotic pressure, we get:
π = (n/V)×RT
π = [(960.76 / M) / 117.8] × 8
Vapour pressure of pure benzene at 353 K, P₁ = 10.0268 × 10⁴ Nm⁻²
Vapour pressure of solution, P₂ = 9.8960 × 10⁴ Nm⁻²
Density of benzene, d = 8.149 × 10² g dm⁻³
To find: Osmotic pressure of the solution
Formula used:
For a dilute solution, the osmotic pressure (π) is given by the equation:
π = (n/V)×RT
where n is the number of moles of the solute, V is the volume of the solution, R is the gas constant and T is the temperature.
The number of moles of the solute can be calculated as:
n = (m/M)
where m is the mass of the solute and M is its molar mass.
Solution:
Molar mass of benzene, C₆H₆ = 6 × 12.01 + 6 × 1.008 = 78.11 g mol⁻¹
Let us assume that we have 1000 g of the solution.
Then, the mass of benzene in the solution = (1000 - m) g
From Raoult's law, we know that:
P₂ = X₂ × P₁
where X₂ is the mole fraction of benzene in the solution.
Mole fraction of benzene in the solution can be calculated as:
X₂ = moles of benzene / total moles
Let the moles of benzene be n₂.
Then, the total moles can be calculated as:
n₁ + n₂ = (mass of benzene / Molar mass of benzene) + (mass of solute / Molar mass of solute)
Since the solute is non-volatile, its vapour pressure is negligible and hence its mole fraction can be assumed to be equal to 1.
Therefore, the above equation can be simplified as:
n₂ = (mass of benzene / Molar mass of benzene)
Thus, we have:
X₂ = n₂ / (n₁ + n₂)
X₂ = (mass of benzene / Molar mass of benzene) / [(mass of benzene / Molar mass of benzene) + (mass of solute / Molar mass of solute)]
Substituting the given values, we get:
9.8960 × 10⁴ = [(1000 - m) / 78.11] / [m / (M × 78.11)] × 10.0268 × 10⁴
Solving for m, we get:
m = 39.24 g
Mass of solute, w = (1000 - m) g = 960.76 g
Number of moles of the solute, n = (w / M) = (960.76 / M) mol
Volume of the solution, V = (w / d) = (960.76 / 8.149) dm³ = 117.8 dm³
Substituting the given values in the equation for osmotic pressure, we get:
π = (n/V)×RT
π = [(960.76 / M) / 117.8] × 8
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The vapour pressure of pure benzene at 353 K is 10.0268 x 104 Nm-2 and that of a solution containing non-volatile solute in benzene is 9.89600 x 104Nm-2. (d of benzene = 8.149 x 102 g dm-3). Thus, osmotic pressure of the solution isa)3.94 x 105 Nm-2b)4.00x 105 Nm-2c)3.94 x 106 Nm-2d)4.00x 106 Nm-2Correct answer is option 'A'. Can you explain this answer?
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The vapour pressure of pure benzene at 353 K is 10.0268 x 104 Nm-2 and that of a solution containing non-volatile solute in benzene is 9.89600 x 104Nm-2. (d of benzene = 8.149 x 102 g dm-3). Thus, osmotic pressure of the solution isa)3.94 x 105 Nm-2b)4.00x 105 Nm-2c)3.94 x 106 Nm-2d)4.00x 106 Nm-2Correct answer is option 'A'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The vapour pressure of pure benzene at 353 K is 10.0268 x 104 Nm-2 and that of a solution containing non-volatile solute in benzene is 9.89600 x 104Nm-2. (d of benzene = 8.149 x 102 g dm-3). Thus, osmotic pressure of the solution isa)3.94 x 105 Nm-2b)4.00x 105 Nm-2c)3.94 x 106 Nm-2d)4.00x 106 Nm-2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The vapour pressure of pure benzene at 353 K is 10.0268 x 104 Nm-2 and that of a solution containing non-volatile solute in benzene is 9.89600 x 104Nm-2. (d of benzene = 8.149 x 102 g dm-3). Thus, osmotic pressure of the solution isa)3.94 x 105 Nm-2b)4.00x 105 Nm-2c)3.94 x 106 Nm-2d)4.00x 106 Nm-2Correct answer is option 'A'. Can you explain this answer?.
The vapour pressure of pure benzene at 353 K is 10.0268 x 104 Nm-2 and that of a solution containing non-volatile solute in benzene is 9.89600 x 104Nm-2. (d of benzene = 8.149 x 102 g dm-3). Thus, osmotic pressure of the solution isa)3.94 x 105 Nm-2b)4.00x 105 Nm-2c)3.94 x 106 Nm-2d)4.00x 106 Nm-2Correct answer is option 'A'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The vapour pressure of pure benzene at 353 K is 10.0268 x 104 Nm-2 and that of a solution containing non-volatile solute in benzene is 9.89600 x 104Nm-2. (d of benzene = 8.149 x 102 g dm-3). Thus, osmotic pressure of the solution isa)3.94 x 105 Nm-2b)4.00x 105 Nm-2c)3.94 x 106 Nm-2d)4.00x 106 Nm-2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The vapour pressure of pure benzene at 353 K is 10.0268 x 104 Nm-2 and that of a solution containing non-volatile solute in benzene is 9.89600 x 104Nm-2. (d of benzene = 8.149 x 102 g dm-3). Thus, osmotic pressure of the solution isa)3.94 x 105 Nm-2b)4.00x 105 Nm-2c)3.94 x 106 Nm-2d)4.00x 106 Nm-2Correct answer is option 'A'. Can you explain this answer?.
Solutions for The vapour pressure of pure benzene at 353 K is 10.0268 x 104 Nm-2 and that of a solution containing non-volatile solute in benzene is 9.89600 x 104Nm-2. (d of benzene = 8.149 x 102 g dm-3). Thus, osmotic pressure of the solution isa)3.94 x 105 Nm-2b)4.00x 105 Nm-2c)3.94 x 106 Nm-2d)4.00x 106 Nm-2Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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