Question

The unit of ebullioscopic constant is:
A)\[\text{K Kg mo}{{\text{l}}^{\text{-1}}}\text{ or K}{{\left( \text{molality} \right)}^{\text{-1}}}\]
B) \[\text{mol Kg }{{\text{K}}^{\text{-1}}}\text{ or }{{\text{K}}^{\text{-1}}}\text{ (molality)}\]
C)$\text{Kg mo}{{\text{l}}^{\text{-1}}}\text{ }{{\text{K}}^{\text{-1}}}\text{ or }{{\text{K}}^{\text{-1}}}\text{ (molality}{{\text{)}}^{\text{-1}}}$
D) $\text{K mo}{{\text{l}}^{\text{-1}}}\text{ Kg or K (molality)}$

Answer 1

Hint: Elevation in the boiling point depends on the molality of the solute (m). This $\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}$ is related to the ebullioscopic constant ${{\text{K}}_{\text{b}}}$and the temperature T as: $\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\text{=}{{\text{K}}_{\text{b}}}\text{m}$.Since ${{\text{K}}_{\text{b}}}$is the constant it has the units of ${{\text{K}}_{\text{b}}}\text{=}\dfrac{\text{T}}{\text{m}}$ .

Complete answer:
The molal boiling point elevation or ebullioscopic constant of the solvent is defined as the elevation in the boiling point of the solution which may be theoretically produced when one mole of the solute is dissolved in 1 $\text{Kg}$ of the solvent. It is denoted by the${{\text{K}}_{\text{b}}}$.
The boiling point, \[{{\text{T}}_{\text{b}}}\] of a liquid, is the temperature at which its vapor pressure is equal to the atmospheric pressure. When a non-volatile solute is added to a liquid, the vapor pressure of the liquid is decreased.
The difference in the boiling point that is an elevation in boiling point $\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}$ . It is related to the heat of vaporization \[\text{ }\!\!\Delta\!\!\text{ }{{\text{H}}_{\text{vap}}}\] . The formula for the elevation in boiling point is as:
\[\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\text{=}\dfrac{\text{RT}_{\text{b}}^{\text{2}}{{\text{M}}_{\text{1}}}\text{m}}{\text{ }\!\!\Delta\!\!\text{ }{{\text{H}}_{\text{vap}}}}\]
Where the factor \[\dfrac{\text{RT}_{\text{b}}^{\text{2}}{{\text{M}}_{\text{1}}}}{\text{ }\!\!\Delta\!\!\text{ }{{\text{H}}_{\text{vap}}}}\] is constant.
Where,
R is the gas constant
\[{{\text{T}}_{\text{b}}}\] Is the boiling point of the solvent
M is the molar mass of the solvent
\[\text{ }\!\!\Delta\!\!\text{ }{{\text{H}}_{\text{vap}}}\] is the molar enthalpy of vaporization
The factor \[\dfrac{\text{RT}_{\text{b}}^{\text{2}}{{\text{M}}_{\text{1}}}}{\text{ }\!\!\Delta\!\!\text{ }{{\text{H}}_{\text{vap}}}}\] is denoted by the ${{\text{K}}_{\text{b}}}$ and called as the molal boiling point elevation constant of the solvent.
Let's find out the unit of ebullioscopic constant.
$\text{ }{{\text{K}}_{\text{b}}}=\dfrac{\text{RT}_{\text{b}}^{\text{2}}{{\text{M}}_{\text{1}}}}{\text{ }\!\!\Delta\!\!\text{ }{{\text{H}}_{\text{vap}}}}\text{ }$
We know the units gas constant R, elevation in boiling point\[{{\text{T}}_{\text{b}}}\], molar mass M and heat of vaporization \[\text{ }\!\!\Delta\!\!\text{ }{{\text{H}}_{\text{vap}}}\] as follows:

Quantity	Symbol	Unit
Gas constant	R	$\text{J}\text{.}{{\text{K}}^{\text{-1}}}\text{.mo}{{\text{l}}^{\text{-1}}}$
Boiling point temperature	\[{{\text{T}}_{\text{b}}}\]	$\text{K}$
Molality of solutions	m	$\text{Kg mo}{{\text{l}}^{\text{-1}}}$
Heat of vaporization	\[\text{ }\!\!\Delta\!\!\text{ }{{\text{H}}_{\text{vap}}}\]	$\text{J}\text{.mo}{{\text{l}}^{\text{-1}}}$

Let's substitute the values of the quantity in the ebullioscopic constant, we have
${{\text{K}}_{\text{b}}}\text{=}\dfrac{\left( \text{J}\text{.}{{\text{K}}^{\text{-1}}}\text{.mo}{{\text{l}}^{\text{-1}}} \right){{\left( \text{K} \right)}^{\text{2}}}\left( \text{Kg mo}{{\text{l}}^{\text{-1}}} \right)}{\left( \text{J}\text{.mo}{{\text{l}}^{\text{-1}}} \right)}$
Cancel out the terms from the numerator and denominator. We have, \[{{\text{K}}_{\text{b}}}\text{=}\dfrac{\left( \text{}\text{.}{{{\text{}}}^{\text{-1}}}\text{.o}{{\text{l}}^{\text{-1}}} \right){{\left( \text{K} \right)}^{{\text{}}}}\left( \text{Kg mo}{{\text{l}}^{\text{-1}}} \right)}{\left( \text{}\text{.o}{{\text{l}}^{\text{-1}}} \right)}\]
Or ${{\text{K}}_{\text{b}}}\text{=}\left( \text{K} \right)\left( \text{Kg mo}{{\text{l}}^{\text{-1}}} \right)$
Thus, the ${{\text{K}}_{\text{b}}}$ has the unit of $\text{K Kg mo}{{\text{l}}^{\text{-1}}}$.

Hence, (D) is the correct option.

Additional information:
Elevation in the boiling point is a colligative property. It means the change in temperature$\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}$, the difference in freezing point $\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{f}}}$is directly related to the number of particles dissolved in the solvent and independent of the nature of the solute.
The depression in freezing point $\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{f}}}$ is related to the cryoscopy constant or the molal freezing point depression constant ${{\text{K}}_{\text{f}}}$has the units of $\text{K Kg mo}{{\text{l}}^{\text{-1}}}$.

Note:
The units of proportionality constant are arranged such that it maintains the units of the left and right-hand side equal. The proportionality constant is arranged in such a way that it cancels out the unit which remains in the equation.