Mark the correct relationship between the boiling points of very dilute solutions of `BaCl_(2)(t_(1))` and `LCl(t_(2))` ,having the same molarity

Options

A. `t_(1)=t_(2)`
B. `t_(1)gt t_(2)`
C. `t_(2)gt t_(1)`
D. `t_(2)` is approximately equal to `t_(1)`

Explanation

To determine the correct relationship between the boiling points of very dilute solutions of BaCl₂ (T₁) and LiCl (T₂) having the same molarity, we can follow these steps:

### Step 1: Understand the boiling point elevation concept
The boiling point elevation (ΔT_b) is given by the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \( \Delta T_b \) = elevation in boiling point
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( K_b \) = ebullioscopic constant of the solvent
- \( m \) = molality of the solution

### Step 2: Identify the van 't Hoff factors
For the two solutes:
- **BaCl₂** dissociates into 3 ions: Ba²⁺ and 2 Cl⁻. Thus, \( i = 3 \).
- **LiCl** dissociates into 2 ions: Li⁺ and Cl⁻. Thus, \( i = 2 \).

### Step 3: Compare the effective concentration (i × m)
Since both solutions have the same molarity (m), we can compare the effective concentration:
- For BaCl₂: \( i \cdot m = 3m \)
- For LiCl: \( i \cdot m = 2m \)

### Step 4: Determine the boiling point elevation
Since the boiling point elevation is directly proportional to \( i \cdot m \):
- The boiling point elevation for BaCl₂ (ΔT_b for BaCl₂) will be greater than that for LiCl (ΔT_b for LiCl) because \( 3m > 2m \).

### Step 5: Conclude the boiling point relationship
Since the boiling point elevation is greater for BaCl₂, it means that the boiling point of the BaCl₂ solution (T₁) will be higher than that of the LiCl solution (T₂):
\[ T_1 > T_2 \]

### Final Answer
The correct relationship is:
\[ T_1 > T_2 \]

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