Raoult's Law
The vapor pressure of pure benzene at 298 K is 100 mm Hg. When 2.0 g of a non-volatile solute is dissolved in 78 g of benzene, the vapor pressure drops to 98 mm Hg. The molar mass of the solute is:
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About This Question
- Subject
- chemistry
- Chapter
- solutions
- Topic
- raoult's law
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
100 g/mol
According to Raoult's Law for relative lowering of vapor pressure: P0P0−Ps=Xsolute≈n1n2
- Values: P0=100 mm Hg, Ps=98 mm Hg.
- Moles of benzene (n1) = molar massmass=7878=1 mol.
- Fractional lowering: 100100−98=0.02.
- Since Xsolute=0.02 and n1=1, we have n1n2≈0.02⇒n2=0.02 mol.
- Molar mass M2=n2mass of solute=0.022.0=100 g/mol.
This medium difficulty chemistry question is from the chapter solutions, covering the topic of raoult's law. It appeared in the 2025 exam.
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