The elevation in boiling point of a solution `dT_(b)` is related with molality of solution `(m)` by the reaction:
`dT_(b)=[(RT_(b)^(2))/(DeltaH_(vap))][(M_(1))/(1+mM_(1))]`, where `M_(1)` is moalr mass of solvent and `DeltaH_(vap)` is heat of vaporisation of solvent. For a dilute solution, the relation `((delT_(b))/(delT_(n)))_(mto0)`
gives:

Options

A. Molal ebullioscopic constant
B. Elevation in boiling point
C. Boiling point of solvent
D. Elevation in boiling point `(DeltaT_(b))` becomes more predominant

Answer

To solve the problem, we need to analyze the given equation for the elevation in boiling point of a solution:

\[ \Delta T_b = \left(\frac{RT_b^2}{\Delta H_{vap}}\right) \left(\frac{M_1}{1 + mM_1}\right) \]

where:
- \(\Delta T_b\) is the elevation in boiling point,
- \(R\) is the universal gas constant,
- \(T_b\) is the boiling point of the solvent,
- \(\Delta H_{vap}\) is the heat of vaporization of the solvent,
- \(M_1\) is the molar mass of the solvent,
- \(m\) is the molality of the solution.

### Step 1: Analyze the equation for dilute solutions
For a dilute solution, the molality \(m\) tends to 0. Thus, we can simplify the term \(1 + mM_1\) in the denominator:

\[ 1 + mM_1 \approx 1 \quad \text{(as \(m \to 0\))} \]

### Step 2: Substitute the simplified term into the equation
Substituting this into the original equation gives:

\[ \Delta T_b \approx \left(\frac{RT_b^2}{\Delta H_{vap}}\right) \left(M_1\right) \]

### Step 3: Differentiate \(\Delta T_b\) with respect to \(T_n\)
To find the relationship as \(m\) approaches 0, we need to differentiate \(\Delta T_b\) with respect to \(T_n\):

\[ \frac{d(\Delta T_b)}{d(T_n)} \quad \text{(as \(m \to 0\))} \]

### Step 4: Evaluate the limit
From the previous steps, we can see that:

\[ \frac{d(\Delta T_b)}{d(T_n)} = \frac{RT_b^2}{\Delta H_{vap}} \cdot M_1 \]

This expression represents a constant, which we can denote as \(K_b\) (the molal ebullioscopic constant).

### Conclusion
Thus, for a dilute solution, the relationship gives us:

\[ \frac{\Delta T_b}{\Delta T_n} \bigg|_{m \to 0} = K_b \]

### Final Answer
The correct interpretation of the relationship as \(m\) approaches 0 is that it gives us the molal ebullioscopic constant \(K_b\).