Question

 If the elevation in the boiling point of a solution of 10 g of solute (mol. wt.= 100) in 100 g of water is \[\text{  }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\text{ }\] , the ebullioscopic constant of water is: 

(A) $\dfrac{\text{  }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\text{ }}{10}$ 

(B) \[\text{  }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\text{ }\]

(C) \[\text{ 10 }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\text{ }\]

(D) \[\text{ 100 }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\text{ }\]

Answer 1

Hint: The elevation of the boiling point $\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}$ is a colligative property. It depends on the amount of solute. The difference in the boiling point is stated as:

$\text{  }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\text{ =  }{{\text{K}}_{\text{b}}}\text{ m  = }{{\text{K}}_{\text{b}}}\times \dfrac{{{\text{w}}_{\text{2}}}}{{{\text{M}}_{\text{2}}}\text{  }\!\!\times\!\!\text{  }{{\text{w}}_{\text{1}}}}$

Where,

 $\text{ }{{\text{K}}_{\text{b}}}\text{ }$ is the ebullioscopic constant

 $\text{ }{{\text{w}}_{\text{2}}}\text{ }$ is the mass of solute 

$\text{ }{{\text{w}}_{1}}\text{ }$is mass of solvent 

$\text{ }{{\text{M}}_{\text{2}}}\text{ }$ is the molar mass of solute

Complete Solution:

The boiling point, $\text{  }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\text{ }$ of a liquid, is the temperature at which the vapour pressure is equal to the atmospheric pressure. When a non-volatile solute is added to a liquid, the vapour pressure of the liquid is decreased. Hence, it must be heated to higher temperatures so that its vapour pressure becomes equal to that of the atmospheric pressure. This means that the addition of a non-volatile solute to a liquid raises its boiling point.

The elevation in boiling point $\text{  }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\text{ }$is related to the molar mass of the solute. The relation between the elevations in boiling point to the molality of solute is stated as follows:

$\text{  }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\text{  =  }{{\text{K}}_{\text{b}}}\text{ m }$

If ${{\text{w}}_{\text{2}}}\text{ kg}$of the solute of the molar mass ${{\text{M}}_{\text{2}}}$ is dissolved in ${{\text{w}}_{1}}\text{ kg}$ of the solvent, then the number of moles of the solute dissolved in $\text{1 kg}$ of the solvent would be given by,

$\text{ m }=\text{ }\dfrac{{{\text{w}}_{\text{2}}}}{{{\text{M}}_{\text{2}}}\text{  }\!\!\times\!\!\text{  }{{\text{w}}_{\text{1}}}}$

Then the equation becomes, 

$\text{  }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\text{ = }{{\text{K}}_{\text{b}}}\times \dfrac{{{\text{w}}_{\text{2}}}}{{{\text{M}}_{\text{2}}}\text{ }\!\!\times\!\!\text{ }{{\text{w}}_{\text{1}}}}$                                ....................  (1)

We are given the following data:

Elevation in boiling point, $\text{  }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\text{ }$

Weight of the compound given,${{w}_{2}}\text{ =  10 g}$

Weight of solvent, ${{w}_{\text{1 }}}=\text{ 100 g}$

The molecular weight of the compound, $\text{M}=\text{ 100 }$

We have to find the Ebullioscopic constant, $\text{ }{{\text{K}}_{\text{b}}}\text{ }$

On rearrangement of equation (1), we have,

$\text{  }{{\text{K}}_{\text{b}}}=\dfrac{\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\times {{\text{M}}_{\text{2}}}\text{ }\!\!\times\!\!\text{ }{{\text{w}}_{\text{1}}}}{{{\text{w}}_{\text{2}}}}\text{ }$

Let's substitute the values in the equation. We have,

$\text{  }{{\text{K}}_{\text{b}}}=\dfrac{\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\times 100\text{  }\!\!\times\!\!\text{ 100}}{10 X 1000}\text{ =  }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}$     (Since we are talking about molalilty)

Therefore, the ebullioscopic constant  ${{\text{K}}_{\text{b}}}$ of the compound is equal to the$\text{  }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}$.

Hence, (B) is the correct option.

Note: If the molality$\text{ }m=1\text{ }$, that is 1 mole of the solute dissolved in the 1 kilogram of the solvent then, $\text{  }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\text{  =  }{{\text{K}}_{\text{b}}}\text{ }$.Thus, molal boiling point elevation constant. The units of ${{\text{K}}_{\text{b}}}$ are $\text{ K kg mol}{{\text{l}}^{-1}}\text{ }$. The units of  ${{\text{K}}_{\text{b}}}$ coming out to be,$\text{ }\dfrac{\left( \text{J }{{\text{K}}^{-1}}\text{mo}{{\text{l}}^{-1}} \right)\left( {{\text{K}}^{2}} \right)\left( \text{kg mo}{{\text{l}}^{-1}} \right)}{\left( \text{J mo}{{\text{l}}^{-1}} \right)}\text{ = K kg mol}{{\text{l}}^{-1}}\text{ }$