Correct Answer - Option 3 : 1 : 5

**Concept:**

**Boiling point elevation formula:**

Let T_{0} denote boiling point of pure liquid, and T_{b} denote the boiling point of solution (solute + pure liquid). Then ΔT_{b} ∝ m

ΔT_{b} = K_{b} × m

Where, \(\Delta {T_b} = T_b^0 - {T_b}\) is elevation in boiling point,

\(m_{b}^{'}\) is the molality,

\(K_{b}^{'}\) is the molal elevation, boling point or ebullioscopic constant whose depends only on the solvent.

Then ΔT_{b }= K_{b×m}

K_{b} = Ebullioscopic constant

m = Molar mass of the solvent

**Calculation:**

In this problem, boiling point elevation formula is used, which is:

ΔT_{b }= K_{b×m}

Now, the ratio of the elevation in their boiling points is:

\(\frac{\text{ }\!\!\Delta\!\!\text{ }{{T}_{b}}\left( A \right)}{\text{ }\!\!\Delta\!\!\text{ }{{T}_{b}}\left( B \right)}=\frac{{{K}_{b}}\left( A \right)\times {{m}_{A}}}{{{K}_{b}}\left( B \right)\times {{m}_{B}}}\)

From question, m_{A} = m_{B}

\(\Rightarrow \frac{\text{ }\!\!\Delta\!\!\text{ }{{T}_{b}}\left( A \right)}{\text{ }\!\!\Delta\!\!\text{ }{{T}_{b}}\left( B \right)}=\frac{{{K}_{b}}\left( A \right)}{{{K}_{b}}\left( B \right)}\)

From the question, the **ebullioscopic constants are in the ratio of 1 : 5**

\(\therefore \frac{\text{ }\!\!\Delta\!\!\text{ }{{T}_{b}}\left( A \right)}{\text{ }\!\!\Delta\!\!\text{ }{{T}_{b}}\left( B \right)}=\frac{1}{5}\)