It is found that elevation in boiling point of a given aq. `NaCl` solution is equal to depression in freezing point of `0.25` molal aq. `Na_(2)CO_(3)` solution. If `K_(f)=1.86^(@) "mol"^(-1) kg`, calculate molality of `NaCl` solution, assume complete ionisation of `NaCl` and `Na_(2)CO_(3)` in aqueous solution.

Explanation

To solve the problem, we need to understand the relationship between the elevation in boiling point and the depression in freezing point, and how they relate to the molality of the solutions involved.

### Step-by-Step Solution:

1. **Understand the given data**:
- We have a solution of NaCl and a solution of Na2CO3.
- The elevation in boiling point of the NaCl solution is equal to the depression in freezing point of the Na2CO3 solution.
- The molality of the Na2CO3 solution is given as 0.25 molal.
- The cryoscopic constant (Kf) is given as 1.86 °C kg/mol.

2. **Determine the van 't Hoff factor (i)**:
- For NaCl, which dissociates into Na⁺ and Cl⁻, the van 't Hoff factor (i) is 2.
- For Na2CO3, which dissociates into 2 Na⁺ and 1 CO3²⁻, the van 't Hoff factor (i) is 3.

3. **Calculate the depression in freezing point (ΔTf) for Na2CO3**:
- The formula for depression in freezing point is given by:
\[ \Delta Tf = i \cdot Kf \cdot m \]
- Substituting the values for Na2CO3:
\[ \Delta Tf = 3 \cdot 1.86 \cdot 0.25 \]
- Calculate ΔTf:
\[ \Delta Tf = 3 \cdot 1.86 \cdot 0.25 = 1.395 \, °C \]

4. **Set up the equation for elevation in boiling point (ΔTb) for NaCl**:
- The elevation in boiling point is given by:
\[ \Delta Tb = i \cdot Kb \cdot m \]
- For NaCl, we have:
\[ \Delta Tb = 2 \cdot Kb \cdot m_{NaCl} \]
- Since ΔTb = ΔTf, we can equate the two:
\[ 2 \cdot Kb \cdot m_{NaCl} = 1.395 \]

5. **Determine Kb for water**:
- The ebullioscopic constant (Kb) for water is approximately 0.52 °C kg/mol.

6. **Substitute Kb into the equation**:
- Now we substitute Kb into the equation:
\[ 2 \cdot 0.52 \cdot m_{NaCl} = 1.395 \]

7. **Solve for m_{NaCl}**:
- Rearranging the equation:
\[ m_{NaCl} = \frac{1.395}{2 \cdot 0.52} \]
- Calculate m_{NaCl}:
\[ m_{NaCl} = \frac{1.395}{1.04} \approx 1.34 \, molal \]

### Final Answer:
The molality of the NaCl solution is approximately **1.34 molal**.