Acids And Bases

Let P⁰ be the vapour pressure of pure water and P_s be the vapour pressure of the solution.
Given:

• Moles of ionic salt (solute), = 0.1 mol
• Mass of water (solvent) = 1.8 kg = 1800 g
• Molar mass of water (H₂O) = 18 g/mol
• Moles of water, = Mass of water / Molar mass of water = 1800 g / 18 g/mol = 100 mol
• Vapour pressure of pure water, P⁰ = 60.000 mm of Hg
• Vapour pressure of the solution, P_s = 59.724 mm of Hg
• Degree of dissociation, α = 90% = 0.90

According to Raoult's law, the relative lowering of vapour pressure is given by: $ is the mole fraction of all solute particles in the solution.
If 'i' is the Van't Hoff factor, then the effective number of moles of solute is .
So, \) $

The final answer is

Identification of Species in Chemical Scheme

The solution involves identifying the species and in the given chemical scheme. The correct answer is:

and

To arrive at this conclusion, let's consider the possible oxidation and reduction reactions that could occur based on the options provided. The presence of suggests the involvement of potassium permanganate, which often acts as an oxidizing agent, converting ions into gas by gaining electrons. This reaction supports the identification of and as and , respectively, given the volatile and reactive nature of these compounds in typical redox reactions.

Step 1: Definition of acidic hydrogen
An acidic hydrogen is one that can be easily removed as a proton ( ), resulting in a stable conjugate base. The more stable the conjugate base, the more acidic the hydrogen.
Step 2: Analyze conjugate base stability
- In Option (B), the removal of the hydrogen leads to a carbanion which is resonance stabilized by conjugation with two adjacent double bonds. This forms an allylic anion extended over a conjugated cyclic system. - In Options (A), (C), and (D), although allylic positions are present, the resonance stabilization is less extensive compared to (B). Especially, (A) and (C) are sterically hindered or less effectively delocalized.
Step 3: Resonance in Option (B)
Removing the proton from the carbon adjacent to two double bonds gives a conjugate base that can delocalize the negative charge over a five-carbon conjugated system, similar to a cyclopentadienyl-like system — which is known for its aromatic stabilization when fully conjugated.
Conclusion: Option (B) has the most stable conjugate base, hence the most acidic hydrogen.
Final Answer: