# 1 g of a non-volatile non-electrolyte solute is dissolved in 100 g of two different solvents A and B whose ebullioscopic constants are in the ratio of

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1 g of a non-volatile non-electrolyte solute is dissolved in 100 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 5. The ratio of the elevation in their boiling points $\frac{{{\rm{\Delta }}{T_b}\left( A \right)}}{{{\rm{\Delta }}{T_b}\left( B \right)}}$, is:
1. 5 : 1
2. 10 : 1
3. 1 : 5
4. 1 : 0.2

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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

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