
1 g of 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 is $\dfrac{\Delta {{T}_{b}}(A)}{\Delta {{T}_{b}}(B)}$ , is:
A. 5 : 1
B. 10 : 1
C. 1 : 5
D. 1 : 0.2
Answer
566.1k+ views
Hint:. The relationship between ebullioscopic constants and elevation in boiling point of the solution is as follows.
\[(\Delta {{T}_{b}})=({{K}_{b}})(m)\]
Where $\Delta {{T}_{b}}$ = boiling temperature
${{K}_{b}}$ = Boiling point of ebullioscopic constant or boiling point
m = molality of the solution
Complete step by step answer:
- We have to find the ratio of elevation in the boiling point of the solution after adding non-volatile non electrolyte solutes to two solvents A and B.
- The amount of solute added in 100 g of solvent A is 1 g.
- The molality of solution A and Solution B is the same because they have the same solute in the same concentration in the same amount (100 g) of the solvents.
- Therefore the solutions A and B have the same molality (m).
- The elevation in boiling points of solutions A and B is as follows.
\[\dfrac{{{(\Delta {{T}_{b}})}_{A}}}{{{(\Delta {{T}_{b}})}_{B}}}=\dfrac{{{({{K}_{b}})}_{A}}}{{{({{K}_{b}})}_{B}}}\times \dfrac{{{m}_{A}}}{{{m}_{B}}}\]
- In the above formula, the molality of both the solutions A and B are the same. So, the molality is going to cancel each other out.
- In the question it is given that the ebullioscopic constants of A and B solutions are in the ratio of 1: 5.
- Therefore
\[\begin{align}
& \dfrac{{{(\Delta {{T}_{b}})}_{A}}}{{{(\Delta {{T}_{b}})}_{B}}}=\dfrac{{{({{K}_{b}})}_{A}}}{{{({{K}_{b}})}_{B}}}\times \dfrac{{{m}_{A}}}{{{m}_{B}}} \\
& \dfrac{{{(\Delta {{T}_{b}})}_{A}}}{{{(\Delta {{T}_{b}})}_{B}}}=\dfrac{1}{5} \\
\end{align}\]
So, the correct answer is “Option C”.
Note: If the molality of the solutions A and B are same then the elevation in boiling point of the solutions will be same as the ebullioscopic constants. If the molality of the solutions is different then the ratio the elevation in boiling point varies.
\[(\Delta {{T}_{b}})=({{K}_{b}})(m)\]
Where $\Delta {{T}_{b}}$ = boiling temperature
${{K}_{b}}$ = Boiling point of ebullioscopic constant or boiling point
m = molality of the solution
Complete step by step answer:
- We have to find the ratio of elevation in the boiling point of the solution after adding non-volatile non electrolyte solutes to two solvents A and B.
- The amount of solute added in 100 g of solvent A is 1 g.
- The molality of solution A and Solution B is the same because they have the same solute in the same concentration in the same amount (100 g) of the solvents.
- Therefore the solutions A and B have the same molality (m).
- The elevation in boiling points of solutions A and B is as follows.
\[\dfrac{{{(\Delta {{T}_{b}})}_{A}}}{{{(\Delta {{T}_{b}})}_{B}}}=\dfrac{{{({{K}_{b}})}_{A}}}{{{({{K}_{b}})}_{B}}}\times \dfrac{{{m}_{A}}}{{{m}_{B}}}\]
- In the above formula, the molality of both the solutions A and B are the same. So, the molality is going to cancel each other out.
- In the question it is given that the ebullioscopic constants of A and B solutions are in the ratio of 1: 5.
- Therefore
\[\begin{align}
& \dfrac{{{(\Delta {{T}_{b}})}_{A}}}{{{(\Delta {{T}_{b}})}_{B}}}=\dfrac{{{({{K}_{b}})}_{A}}}{{{({{K}_{b}})}_{B}}}\times \dfrac{{{m}_{A}}}{{{m}_{B}}} \\
& \dfrac{{{(\Delta {{T}_{b}})}_{A}}}{{{(\Delta {{T}_{b}})}_{B}}}=\dfrac{1}{5} \\
\end{align}\]
So, the correct answer is “Option C”.
Note: If the molality of the solutions A and B are same then the elevation in boiling point of the solutions will be same as the ebullioscopic constants. If the molality of the solutions is different then the ratio the elevation in boiling point varies.
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