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# Calculate the molal elevation constant, ${{K}_{b}}$for water and the boiling of 0.1 molal urea solution. Latent heat of vapourization of water 9.72 $\dfrac{kcal}{mol}$at 373.15 K.(A) $~{{K}_{b}}\text{ }=\text{ }0.512kgmol{{K}^{-1}}\text{ },\text{ }{{T}_{b}}\text{ }=\text{ }373.20K$(B) ${{K}_{b}}\text{ }=\text{ }05.12kgmol{{K}^{-1}},\text{ }{{T}_{b}}\text{ }=\text{ }378.20K$(C) ${{K}_{b}}=\text{ }1.02kgmol{{K}^{-1}}\text{ },\text{ }{{T}_{b}}\text{ }=\text{ }383.20K$(D) ${{K}_{b}}\text{ }=\text{ }0.512kgmol{{K}^{-1}}\text{ },\text{ }{{T}_{b}}\text{ }=\text{ }385.70K$

Last updated date: 13th Jun 2024
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Hint: To calculate this question, we should know about molal elevation constant. We should know that molal elevation constant is defined as the elevation in boiling point produced when solute is introduced in solvent.

> So, first of all we should know about the molal elevation constant. We should define molal elevation constant as the elevation in boiling point produced when one mole of solute is dissolved in 1 kilogram that is 1000 g of the solvent.
> We should know that the unit of molal elevation constant is given as$Kkgmo{{l}^{-1}}$.
> If we do the experiment, we will find that the elevation of the boiling point ($\Delta {{T}_{b}})~$for dilute solutions is directly proportional to the molal concentration of solute in the solution. Therefore, we can write:
$\Delta {{T}_{b}}\propto m$
$\Delta {{T}_{b}}={{K}_{b}}m$
> Here, we should know that m is molality or the number of moles of the solute dissolved in a kg of solvent. ${{K}_{b}}$is the molal elevation constant.
To calculate molal elevation constant(${{K}_{b}}$)we have the following formula:
${{K}_{b}}=\dfrac{R\times {{M}_{1}}\times {{T}_{b}}}{1000\times {{\Delta }_{vap}}H}$
${{M}_{1}}=18gram$
Boiling point of water $\left( {{T}_{b}} \right)=373.15K$
\begin{align} & {{\Delta }_{vap}}H=9.72\dfrac{kcal}{mol}=40629.6\dfrac{J}{mol} \\ & \\ & {{K}_{b}}=0.512kgmol{{K}^{-1}} \\ \end{align}
Putting the value in the above formula we will find the Kb.
\begin{align} & Kb=\dfrac{8.314\times 18\times 373.15}{1000\times 40629.6} \\ & Kb=0.512kgmol{{K}^{-1}} \\ \end{align}
Now, we know that$\Delta {{T}_{b}}={{K}_{b}}m$:
So, by putting the value of molal elevation constant and molality in above formula we will get:
\begin{align} & \Delta {{T}_{b}}=0.512\times 0.1=0.0512K \\ & {{T}_{b}}=373.15+0.0512=373.20K \\ \end{align}
So, from the above discussion and calculation we now know our answer that is option A. $~{{K}_{b}}\text{ }=\text{ }0.512kgmol{{K}^{-1}}\text{ },\text{ }{{T}_{b}}\text{ }=\text{ }373.20K$

Note:
There is a molal depression constant also. We should know that molal depression constant is defined as the depression in freezing point when the molality of the solution is unity, that is one mole of the solute is dissolved in 1000 grams or 1 kilogram of the solvent.