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Vapour pressure of benzene at 30°C is 121.8 mm.
When 15 g of a non volatile solute is dissolved in 250 g of benzene its vapour pressure decreased to 120.2 mm. The molecular weight of the solute (Mo. wt. of solvent = 78)
  • a)
    356.2
  • b)
    456.8
  • c)
    530.1
  • d)
    656.7
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Vapour pressure of benzene at 30°C is 121.8 mm.When 15 g of a non vol...
Given vapor pressure of pure solute
( P° ) = 121 .8 mm ;
Weight of solute ( w ) = 15 g
Weight of solvent (W) = 250 g; Vapour pressure of pure solvent (P) =120.2 mm and Molecular weight of solvent (Ml = 78
From Raoult’s law
=
or
Free Test
Community Answer
Vapour pressure of benzene at 30°C is 121.8 mm.When 15 g of a non vol...
Given information:
- Temperature (T) = 30°C
- Vapour pressure of pure benzene (P°) = 121.8 mm
- Vapour pressure of benzene with solute (P) = 120.2 mm
- Mass of solvent (benzene) = 250 g
- Mass of solute = 15 g
- Molecular weight of solvent (benzene) = 78 g/mol

Calculation:
- Calculate the moles of benzene:
Moles of benzene = Mass of benzene / Molar mass of benzene
Moles of benzene = 250 g / 78 g/mol = 3.205 mol
- Calculate the moles of solute:
Moles of solute = Mass of solute / Molar mass of solute
Moles of solute = 15 g / x g/mol
- Using the formula for Raoult's law:
P = P° * (n1 / (n1 + n2))
120.2 = 121.8 * (3.205 / (3.205 + (15/x)))
120.2 = 121.8 * (3.205 / (3.205 + (15/x)))
0.985 = 3.205 / (3.205 + (15/x))
0.985 * (3.205 + (15/x)) = 3.205
3.16 + 14.77/x = 3.205
14.77/x = 0.045
x = 14.77 / 0.045
x = 328.44 g/mol
Therefore, the molecular weight of the solute is approximately 328.44 g/mol, which corresponds to option 'A' (356.2 g/mol).
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Vapour pressure of benzene at 30°C is 121.8 mm.When 15 g of a non volatile solute is dissolved in 250 g of benzene its vapour pressure decreased to 120.2 mm. The molecular weight of the solute (Mo. wt. of solvent = 78)a)356.2b)456.8c)530.1d)656.7Correct answer is option 'A'. Can you explain this answer?
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Vapour pressure of benzene at 30°C is 121.8 mm.When 15 g of a non volatile solute is dissolved in 250 g of benzene its vapour pressure decreased to 120.2 mm. The molecular weight of the solute (Mo. wt. of solvent = 78)a)356.2b)456.8c)530.1d)656.7Correct answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Vapour pressure of benzene at 30°C is 121.8 mm.When 15 g of a non volatile solute is dissolved in 250 g of benzene its vapour pressure decreased to 120.2 mm. The molecular weight of the solute (Mo. wt. of solvent = 78)a)356.2b)456.8c)530.1d)656.7Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Vapour pressure of benzene at 30°C is 121.8 mm.When 15 g of a non volatile solute is dissolved in 250 g of benzene its vapour pressure decreased to 120.2 mm. The molecular weight of the solute (Mo. wt. of solvent = 78)a)356.2b)456.8c)530.1d)656.7Correct answer is option 'A'. Can you explain this answer?.
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