Correct Answer - Option 3 : 1 : 5
Concept:
Boiling point elevation formula:
Let T0 denote boiling point of pure liquid, and Tb denote the boiling point of solution (solute + pure liquid). Then ΔTb ∝ m
ΔTb = Kb × 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 ΔTb = Kb×m
Kb = Ebullioscopic constant
m = Molar mass of the solvent
Calculation:
In this problem, boiling point elevation formula is used, which is:
ΔTb = Kb×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, mA = mB
\(\Rightarrow \frac{\text{ }\!\!\Delta\!\!\text{ }{{T}_{b}}\left( A \right)}{\text{ }\!\!\Delta\!\!\text{ }{{T}_{b}}\left( B \right)}=\frac{{{K}_{b}}\left( A \right)}{{{K}_{b}}\left( B \right)}\)
From the question, the ebullioscopic constants are in the ratio of 1 : 5
\(\therefore \frac{\text{ }\!\!\Delta\!\!\text{ }{{T}_{b}}\left( A \right)}{\text{ }\!\!\Delta\!\!\text{ }{{T}_{b}}\left( B \right)}=\frac{1}{5}\)