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The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile, non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. Molecular weight of the non-electrolyte is
- a)65.25 g mol-1
- b)60.08 g mol-1
- c)72.08 g mol-1
- d)92.05 g mol-1
Correct answer is option 'A'. Can you explain this answer?
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The vapour pressure of pure benzene at a certain temperature is 640 mm...
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The vapour pressure of pure benzene at a certain temperature is 640 mm...
Given:
Vapour pressure of pure benzene (P1) = 640 mm Hg
Weight of non-volatile, non-electrolyte solid (W2) = 2.175 g
Weight of benzene (W1) = 39.0 g
Vapour pressure of the solution (P2) = 600 mm Hg
To find:
Molecular weight of the non-electrolyte (M2)
Formula Used:
Raoult's Law: P2 = P1*X1
Where P2 is the vapour pressure of the solution, P1 is the vapour pressure of the pure solvent, and X1 is the mole fraction of the solvent in the solution.
Solution:
Step 1: Calculate the mole fraction of benzene in the solution
Mole fraction of benzene (X1) = (moles of benzene)/(total moles)
To calculate the moles of benzene, we need to know the number of moles of benzene and the number of moles of the non-electrolyte.
Step 2: Calculate the moles of benzene
Moles of benzene (n1) = (weight of benzene)/(molar mass of benzene)
The molar mass of benzene (C6H6) = 12.01 g/mol (carbon) + 1.008 g/mol (hydrogen) x 6 = 78.11 g/mol
Plugging in the values:
n1 = 39.0 g / 78.11 g/mol = 0.499 mol
Step 3: Calculate the moles of the non-electrolyte
Moles of the non-electrolyte (n2) = (weight of the non-electrolyte)/(molar mass of the non-electrolyte)
Plugging in the values:
n2 = 2.175 g / M2 g/mol
Step 4: Calculate the mole fraction of benzene
X1 = n1/(n1 + n2)
Substituting the values:
X1 = 0.499 / (0.499 + 2.175/M2)
Step 5: Apply Raoult's Law to find the molecular weight of the non-electrolyte
P2 = P1 * X1
Substituting the values:
600 mm Hg = 640 mm Hg * (0.499 / (0.499 + 2.175/M2))
Simplifying the equation:
600 = 640 * (0.499 / (0.499 + 2.175/M2))
To solve this equation, we can cross-multiply and solve for M2:
600 * (0.499 + 2.175/M2) = 640 * 0.499
299.4 + 1305/M2 = 319.36
1305/M2 = 319.36 - 299.4
1305/M2 = 19.96
M2 = 1305/19.96
M2 = 65.25 g/mol
Therefore, the molecular weight of the non-electrolyte is 65.25 g/mol
Vapour pressure of pure benzene (P1) = 640 mm Hg
Weight of non-volatile, non-electrolyte solid (W2) = 2.175 g
Weight of benzene (W1) = 39.0 g
Vapour pressure of the solution (P2) = 600 mm Hg
To find:
Molecular weight of the non-electrolyte (M2)
Formula Used:
Raoult's Law: P2 = P1*X1
Where P2 is the vapour pressure of the solution, P1 is the vapour pressure of the pure solvent, and X1 is the mole fraction of the solvent in the solution.
Solution:
Step 1: Calculate the mole fraction of benzene in the solution
Mole fraction of benzene (X1) = (moles of benzene)/(total moles)
To calculate the moles of benzene, we need to know the number of moles of benzene and the number of moles of the non-electrolyte.
Step 2: Calculate the moles of benzene
Moles of benzene (n1) = (weight of benzene)/(molar mass of benzene)
The molar mass of benzene (C6H6) = 12.01 g/mol (carbon) + 1.008 g/mol (hydrogen) x 6 = 78.11 g/mol
Plugging in the values:
n1 = 39.0 g / 78.11 g/mol = 0.499 mol
Step 3: Calculate the moles of the non-electrolyte
Moles of the non-electrolyte (n2) = (weight of the non-electrolyte)/(molar mass of the non-electrolyte)
Plugging in the values:
n2 = 2.175 g / M2 g/mol
Step 4: Calculate the mole fraction of benzene
X1 = n1/(n1 + n2)
Substituting the values:
X1 = 0.499 / (0.499 + 2.175/M2)
Step 5: Apply Raoult's Law to find the molecular weight of the non-electrolyte
P2 = P1 * X1
Substituting the values:
600 mm Hg = 640 mm Hg * (0.499 / (0.499 + 2.175/M2))
Simplifying the equation:
600 = 640 * (0.499 / (0.499 + 2.175/M2))
To solve this equation, we can cross-multiply and solve for M2:
600 * (0.499 + 2.175/M2) = 640 * 0.499
299.4 + 1305/M2 = 319.36
1305/M2 = 319.36 - 299.4
1305/M2 = 19.96
M2 = 1305/19.96
M2 = 65.25 g/mol
Therefore, the molecular weight of the non-electrolyte is 65.25 g/mol
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The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile, non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. Molecular weight of the non-electrolyte isa)65.25 g mol-1b)60.08g mol-1c)72.08g mol-1d)92.05g mol-1Correct answer is option 'A'. Can you explain this answer?
Question Description
The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile, non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. Molecular weight of the non-electrolyte isa)65.25 g mol-1b)60.08g mol-1c)72.08g mol-1d)92.05g mol-1Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile, non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. Molecular weight of the non-electrolyte isa)65.25 g mol-1b)60.08g mol-1c)72.08g mol-1d)92.05g mol-1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile, non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. Molecular weight of the non-electrolyte isa)65.25 g mol-1b)60.08g mol-1c)72.08g mol-1d)92.05g mol-1Correct answer is option 'A'. Can you explain this answer?.
The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile, non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. Molecular weight of the non-electrolyte isa)65.25 g mol-1b)60.08g mol-1c)72.08g mol-1d)92.05g mol-1Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile, non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. Molecular weight of the non-electrolyte isa)65.25 g mol-1b)60.08g mol-1c)72.08g mol-1d)92.05g mol-1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile, non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. Molecular weight of the non-electrolyte isa)65.25 g mol-1b)60.08g mol-1c)72.08g mol-1d)92.05g mol-1Correct answer is option 'A'. Can you explain this answer?.
Solutions for The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile, non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. Molecular weight of the non-electrolyte isa)65.25 g mol-1b)60.08g mol-1c)72.08g mol-1d)92.05g mol-1Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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