
The rise in the boiling point of a solution containing 1.8g of glucose in 100g of solvent is ${{0.1}^{\circ }}C$. The molal elevation constant of the liquid is :
A. ${{0.01}^{\circ }}C{{m}^{-1}}$
B. ${{0.1}^{\circ }}C{{m}^{-1}}$
C. ${{1}^{\circ }}C{{m}^{-1}}$
D. ${{10}^{\circ }}C{{m}^{-1}}$
Answer
305.7k+ views
Hint: Molal elevation constant is defined as the elevation in the boiling point when one mole of non- volatile solute is dissolved in the 1 Kg of the solvent. It can also be defined as the ebullioscopic constant. By putting the given values in the formula of molal elevation constant, we are able to find the value.
Formula Used:
Formula to find the molal elevation constant of the liquid is :
$\Delta {{T}_{b}}={{K}_{b}}M$
Where ${{K}_{b}}$= boiling point elevation constant
$\Delta {{T}_{b}}$= elevation in boiling point
And units of ${{K}_{b}}=\dfrac{K-{{K}_{g}}}{mole}$
Complete Step by Step Solution:
Given $\Delta {{T}_{b}}$= ${{0.1}^{\circ }}C$
And m = 1.8g
We have to find the molal constant elevation constant of the liquid.
We know $\Delta {{T}_{b}}={{K}_{b}}M$ …(1)
where ${{K}_{b}}$ = boiling point elevation constant
$\Delta {{T}_{b}}$ = elevation in boiling point
From equation (1),
${{K}_{b}}=\dfrac{\Delta {{T}_{b}}}{m}$
Putting the values in the above equation, we get
${{K}_{b}}=\dfrac{0.1\times 100}{\dfrac{1.8}{180}\times 1000}$
Solving further, we get
${{K}_{b}}=\dfrac{0.1\times 100\times 180}{1.8\times 1000}$
Hence \[{{K}_{b}}\]= ${{1}^{\circ }}C{{m}^{-1}}$
Thus the molal elevation constant of the liquid is ${{1}^{\circ }}C{{m}^{-1}}$
Thus, Option (C) is correct.
Note: If ${{T}^{\circ }}_{b}$ is the boiling point of pure solvent and ${{T}_{b}}$is the boiling point of the solution then the elevation in boiling point is
$\Delta {{T}_{b}}={{T}_{b}}-{{T}_{b}}^{\circ }$
It shows that there exists a relation between the elevation in boiling point and the molality ‘m’ of the solute present in the solution.
$\Delta {{T}_{b}}\propto m$
Then $\Delta {{T}_{b}}={{K}_{b}}m$
Where \[{{K}_{b}}\] is the molal elevation constant or ebullioscopic constant.
Formula Used:
Formula to find the molal elevation constant of the liquid is :
$\Delta {{T}_{b}}={{K}_{b}}M$
Where ${{K}_{b}}$= boiling point elevation constant
$\Delta {{T}_{b}}$= elevation in boiling point
And units of ${{K}_{b}}=\dfrac{K-{{K}_{g}}}{mole}$
Complete Step by Step Solution:
Given $\Delta {{T}_{b}}$= ${{0.1}^{\circ }}C$
And m = 1.8g
We have to find the molal constant elevation constant of the liquid.
We know $\Delta {{T}_{b}}={{K}_{b}}M$ …(1)
where ${{K}_{b}}$ = boiling point elevation constant
$\Delta {{T}_{b}}$ = elevation in boiling point
From equation (1),
${{K}_{b}}=\dfrac{\Delta {{T}_{b}}}{m}$
Putting the values in the above equation, we get
${{K}_{b}}=\dfrac{0.1\times 100}{\dfrac{1.8}{180}\times 1000}$
Solving further, we get
${{K}_{b}}=\dfrac{0.1\times 100\times 180}{1.8\times 1000}$
Hence \[{{K}_{b}}\]= ${{1}^{\circ }}C{{m}^{-1}}$
Thus the molal elevation constant of the liquid is ${{1}^{\circ }}C{{m}^{-1}}$
Thus, Option (C) is correct.
Note: If ${{T}^{\circ }}_{b}$ is the boiling point of pure solvent and ${{T}_{b}}$is the boiling point of the solution then the elevation in boiling point is
$\Delta {{T}_{b}}={{T}_{b}}-{{T}_{b}}^{\circ }$
It shows that there exists a relation between the elevation in boiling point and the molality ‘m’ of the solute present in the solution.
$\Delta {{T}_{b}}\propto m$
Then $\Delta {{T}_{b}}={{K}_{b}}m$
Where \[{{K}_{b}}\] is the molal elevation constant or ebullioscopic constant.
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