Question

The unit of ebullioscopic constant is:
[A]$Kkgmo{{l}^{-1}}orK{{(molality)}^{-1}}$
[B]$molkg{{K}^{-1}}or{{K}^{-1}}(molality)$
[C]$kgmo{{l}^{-1}}{{K}^{-1}}or{{K}^{-1}}{{(molality)}^{-1}}$
[D]$KmolK{{g}^{-1}}orK(molality)$

Answer 1

Hint: Ebullioscopic constant relates the molality and the elevation in boiling point. It is the ratio of boiling point elevation and molality.

Complete answer:
In thermodynamics, we study ebullioscopic constants to find a relation between molality and the elevation in boiling point. The relation is written as-
     \[\Delta T=i{{K}_{b}}b\]
Where, ‘b’ is the elevation in boiling point,
‘i’ is the Van’t Hoff factor which is the number of splitting or forming of the particles of the solute.
As we know that elevation in boiling point is a colligative property therefore, $\Delta T$ does not depend on the nature of the particles rather, it depends on the number of particles that dissolves.
There is a formula which we can use for the calculation of the ebullioscopic constant. In the same formula if we put the units of the terms used, we can find out the unit of the ebullioscopic constant ${{K}_{b}}$. The formula is-
     \[{{K}_{b}}=\dfrac{RT_{b}^{2}M}{\Delta {{H}_{vap}}}\]
Here, R is the universal gas constant and its value is fixed which is 8.314 J/mol.K
M is the molar mass of the solvent and its unit is kg/mol
${{T}_{b}}$ is the boiling point of the solvent in kelvin
And lastly, $\Delta {{H}_{vap}}$ is the molar enthalpy of evaporation and its unit is KJ/mol
Therefore, to find its unit, we can write that\[{{K}_{b}}=\dfrac{Jmo{{l}^{-1}}{{K}^{-1}}\times {{K}^{2}}\times kgmo{{l}^{-1}}}{KJmo{{l}^{-1}}}=Kkgmo{{l}^{-1}}\].
As we can define molality as kg/mol, we can write the unit of ${{K}_{b}}$ as $K{{(molality)}^{-1}}$.
As we can see from the above calculation, the unit of ebullioscopic constant came out to be $Kkgmo{{l}^{-1}}$.

Therefore, the correct answer is option [A] $Kkgmo{{l}^{-1}}orK{{(molality)}^{-1}}$.

Note:
As ebullioscopic constant give us a relation between the molality and the elevation in boiling point, similarly there is a cryoscopic constant which gives us a relation between the molality and the depression in freezing point and it is denoted as ${{K}_{f}}$. Through the process of ebullioscopy and cryoscopy, we can calculate the value of unknown molar mass through a known constant.