Question

At 290 K, a vessel (I) contains equal moles of three liquids (A, B, C). The boiling points of A, B, and C are 348 K, 378 K, and 368 K respectively. Vessel (I) is heated to 300 K and vapors were collected into vessel (II). Identify the correct statements. (Assume vessel (I) contains liquids and vapors and vessel (II) contains only vapors.)
Statements:
I.Vessel – I is rich in liquid
II.Vessel – II is rich in vapors of C.
III. The vapor pressures of A, B, and C in Vessel (I) at 300 K follows the order \( C>A>B \).

• A. \(\text{I, III} \)
• B. \(\text{I, III} \)
• C. \(\text{II, III only} \)
• D. \(\text{I, II, III}\)

Hint:

The component with the lowest boiling point will always contribute more to the vapour phase, while the component with the highest boiling point will remain mostly in the liquid phase.

Answer 1

The Correct Option is A

Step 1: Understanding the Boiling Points The boiling points of the substances determine their volatility. Since C has the lowest boiling point (308 K), it is more volatile and will have a higher proportion in the vapour phase. B has the highest boiling point (373 K) and will remain mostly in the liquid phase. 

Step 2: Distribution of Liquids and Vapours - Vessel (I) retains more of the less volatile component B in the liquid state, making it rich in liquid B.
- Vessel (II) collects more of the most volatile component C, making it rich in vapour of C. 

Step 3: Comparing Vapour Pressures Since vapour pressure is inversely related to boiling point, the order of vapour pressures at 290 K will be: C>A>B.

 Thus, all three statements (I, II, III) are correct.