Answer
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Hint: The molal elevation constant can be given as the division of elevation in boiling point to the molality. As we know that elevation in boiling point has units of temperature and molality has units of $mol/Kg$. So, we can find the units of molal elevation constant from it.
Complete step by step answer:
As we know that the vapour pressure of a solution of non-volatile solute is always less than the vapour pressure of the pure solvent. This is because when the surface is occupied by non-volatile solute it results in less solvent being able to evaporate. Therefore, at the boiling point of pure solvent, the non-volatile solvent will not evaporate or boil. Thus, to boil the non-volatile solution, we need to increases its vapour pressure or temperature. The increased temperature can be shown as:
$\Delta {T_b} = T - T^\circ $ ($T > T^\circ $)
Where $T = $ boiling point of the solution
$T^\circ = $ Boiling point of the pure solvent
$\Delta {T_b} = $ It is the elevation in the boiling point and it is the difference between boiling point of the solution and the boiling point of the pure solvent.
Now, is has been proved experimentally that for a dilute solution the elevation of boiling point (${T_b}$) is directly proportional to the molality of the solution($m$). Therefore, it can be written as:
${T_b}{\text{ }}\alpha {\text{ m}}$
${T_b} = {k_b} \times m$ $ - (1)$
Where $m = $ molality of the solution
${k_b} = $ Molal elevation constant .
Here, molal elevation constant is defined as the elevation in boiling point when the molality of the solute is unity that is when $1{\text{ mole}}$ of the solute is dissolved in $1{\text{ Kg}}$ of the solvent. Hence, from equation $ - (1)$, the molal elevation constant (${k_b}$) can be given as:
${k_b} = \dfrac{{{T_b}}}{m}$
Here, temperature has the units of Kelvin ($K$) and molality can be expressed as $mol/Kg$.
Now, The units of ${k_b} = \dfrac{K}{{mol/Kg}}$
Hence, units of molal elevation constant (${k_b}$) can be given as : $K - Kg/mol$.
Note:
Remember that molal elevation constant (${k_b}$) is related to the elevation in the boiling point and the molal depression constant (${k_f}$) is related to the depression of the freezing point of the solution.
Complete step by step answer:
As we know that the vapour pressure of a solution of non-volatile solute is always less than the vapour pressure of the pure solvent. This is because when the surface is occupied by non-volatile solute it results in less solvent being able to evaporate. Therefore, at the boiling point of pure solvent, the non-volatile solvent will not evaporate or boil. Thus, to boil the non-volatile solution, we need to increases its vapour pressure or temperature. The increased temperature can be shown as:
$\Delta {T_b} = T - T^\circ $ ($T > T^\circ $)
Where $T = $ boiling point of the solution
$T^\circ = $ Boiling point of the pure solvent
$\Delta {T_b} = $ It is the elevation in the boiling point and it is the difference between boiling point of the solution and the boiling point of the pure solvent.
Now, is has been proved experimentally that for a dilute solution the elevation of boiling point (${T_b}$) is directly proportional to the molality of the solution($m$). Therefore, it can be written as:
${T_b}{\text{ }}\alpha {\text{ m}}$
${T_b} = {k_b} \times m$ $ - (1)$
Where $m = $ molality of the solution
${k_b} = $ Molal elevation constant .
Here, molal elevation constant is defined as the elevation in boiling point when the molality of the solute is unity that is when $1{\text{ mole}}$ of the solute is dissolved in $1{\text{ Kg}}$ of the solvent. Hence, from equation $ - (1)$, the molal elevation constant (${k_b}$) can be given as:
${k_b} = \dfrac{{{T_b}}}{m}$
Here, temperature has the units of Kelvin ($K$) and molality can be expressed as $mol/Kg$.
Now, The units of ${k_b} = \dfrac{K}{{mol/Kg}}$
Hence, units of molal elevation constant (${k_b}$) can be given as : $K - Kg/mol$.
Note:
Remember that molal elevation constant (${k_b}$) is related to the elevation in the boiling point and the molal depression constant (${k_f}$) is related to the depression of the freezing point of the solution.
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