Match the entries listed in Column I with appropriate listed in Column II.
`{:(,"Column I",,"Column II"),((A),0.1 M BaCl_(2)" solution",(p),271 K),((B),0.1 M NaCl" solution",(q),270 K),((C ),0.1 M K_(3)[Fe(CN)_(6)]" solution",(r ),268 K),((D),0.1 M Al_(2)(SO_(4))_(3)" solution",(s),269 K):}`
Given : Freezing point of 0.1 M sucrose solution = 272 K

Explanation

To solve the matching question between Column I and Column II, we will use the concept of freezing point depression, which is given by the formula:

\[ \Delta T_f = i \cdot K_f \cdot m \]

Where:
- \(\Delta T_f\) is the depression in freezing point,
- \(i\) is the van 't Hoff factor (number of particles the solute breaks into),
- \(K_f\) is the cryoscopic constant (which we can ignore since we are comparing relative values),
- \(m\) is the molality of the solution.

Given that the freezing point of a 0.1 M sucrose solution is 272 K, we can calculate the depression in freezing point for each solution in Column I.

### Step 1: Calculate the freezing point depression for sucrose
- Freezing point of pure solvent (water) = 273 K
- Freezing point of 0.1 M sucrose = 272 K
- \(\Delta T_f\) for sucrose = \(273 - 272 = 1 K\)

### Step 2: Analyze each solution in Column I

#### A. 0.1 M BaCl₂
- BaCl₂ dissociates into 3 ions: Ba²⁺ + 2Cl⁻ (i = 3)
- \(\Delta T_f = i \cdot m = 3 \cdot 0.1 = 0.3 K\)
- Freezing point = \(273 - 0.3 = 272.7 K\) (not matching)

#### B. 0.1 M NaCl
- NaCl dissociates into 2 ions: Na⁺ + Cl⁻ (i = 2)
- \(\Delta T_f = 2 \cdot 0.1 = 0.2 K\)
- Freezing point = \(273 - 0.2 = 272.8 K\) (not matching)

#### C. 0.1 M K₃[Fe(CN)₆]
- K₃[Fe(CN)₆] dissociates into 4 ions: 3K⁺ + [Fe(CN)₆]³⁻ (i = 4)
- \(\Delta T_f = 4 \cdot 0.1 = 0.4 K\)
- Freezing point = \(273 - 0.4 = 272.6 K\) (not matching)

#### D. 0.1 M Al₂(SO₄)₃
- Al₂(SO₄)₃ dissociates into 5 ions: 2Al³⁺ + 3SO₄²⁻ (i = 5)
- \(\Delta T_f = 5 \cdot 0.1 = 0.5 K\)
- Freezing point = \(273 - 0.5 = 272.5 K\) (not matching)

### Step 3: Compare with Column II
- We know that the freezing point of the sucrose solution is 272 K. Therefore:
- The solution with a freezing point of 271 K must have a depression of 2 K.
- The solution with a freezing point of 270 K must have a depression of 3 K.
- The solution with a freezing point of 269 K must have a depression of 4 K.
- The solution with a freezing point of 268 K must have a depression of 5 K.

### Final Matching
- **A (0.1 M BaCl₂)** matches with **Q (270 K)**
- **B (0.1 M NaCl)** matches with **P (271 K)**
- **C (0.1 M K₃[Fe(CN)₆])** matches with **S (269 K)**
- **D (0.1 M Al₂(SO₄)₃)** matches with **R (268 K)**

### Summary of Matches:
- A → Q
- B → P
- C → S
- D → R