Question

A solution of glucose $ \left[ \text{M}\cdot \text{M=180 g mo}{{\text{l}}^{-1}} \right] $ in water has a boiling point of same solution metal constants for water $ {{\text{K}}_{\text{s}}} $ and $ {{\text{K}}_{\text{b}}} $ are $ 1\cdot 86\text{ kg mo}{{\text{l}}^{-1}} $ and $ 0\cdot 512\text{ K kg} $ most respectively.

Answer 1

Hint: Freezing Point: Freezing point of a substance is defined as the temperature at which the vapour pressure of its liquid is equal to the vapour pressure of corresponding solid.
The difference between the freezing points of pure solvent and its solution is called depression in freezing point.
Depression in freezing point:
 $ \Delta {{\text{T}}_{\text{f}}} $ =Freezing point of the solvent $ - $ Freezing point of solution.
 $ \Delta {{\text{T}}_{\text{f}}} $
Where $ {{\text{K}}_{\text{f}}} $ cryoscopy constant and m is molality of solution

Complete step by step solution
Here the solution of glucose is prepared by adding glucose into water.
So that, Boiling point of pure solvent [water] $ =100{}^\circ \text{ C} $
Boiling point of solution $ =109\cdot 20{}^\circ \text{ C} $
Elevation in Boiling point=Boiling point of solution $ - $ Boiling point of
 $ \Delta {{\text{T}}_{\text{f}}} $
Where $ {{\text{K}}_{\text{b}}} $ =Ebullioscopic constant
m=Molality of solution
 $ \Delta {{\text{T}}_{\text{f}}} $ ………. (1)
Where $ {{\text{K}}_{\text{b}}} $ (Given) = $ 0\cdot 512 $
Put the value of $ \Delta {{\text{T}}_{\text{f}}} $ & $ {{\text{K}}_{\text{b}}} $ in equation (1) and we get value of ‘m’
 $ \text{m}=\dfrac{\Delta \text{T}}{{{\text{K}}_{\text{b}}}} $
 $ \text{m}=\dfrac{9\cdot 20}{0\cdot 512} $
After solving we get m= $ 17\cdot 968\cong 17\cdot 97 $
Now freezing point of pure solvent [Water $ 0{}^\circ $ ]
 $ {{\text{K}}_{\text{f}}} $ Of water given $ =1\cdot 86\text{ k/m} $
 $ \Delta {{\text{T}}_{\text{f}}} $
Put the value of $ {{\text{T}}_{\text{f}}}\text{ }{{\text{K}}_{\text{b}}} $ and m we get the value $ {{\text{T}}_{\text{S}}} $
 $ \begin{align}
  & {{\text{T}}_{\text{s}}}={{\text{T}}_{\text{f}}}-{{\text{K}}_{\text{b}}}\text{m} \\
 & \text{ =0}{}^\circ \text{ C}-1\cdot 86\times 17\cdot 97{}^\circ \text{ C} \\
 & \text{ }=0-33\cdot 42{}^\circ \text{ C} \\
 & \text{ }=-33\cdot 42{}^\circ \text{ C} \\
\end{align} $
 $ \because $ Freezing point of the solution
 $ {{\text{T}}_{\text{s}}}=-33\cdot 42{}^\circ \text{ C} $ .

Note
Anti freezing solutions: Water is used in the radiators of cars and other automobiles.
In some countries, where atmospheric temperature less than zero, the water present in radiator freezes, in these condition anti freezing solution are used
Anti-freezing solution is prepared by dissolving ethylene glycol in water. Freezing points can be lowered to the desired extent by varying the concentration of ethylene glycol.
Freezing mixture: It is a mixture of ice and common salt (NaCl). It is used in the making of ice cream and in the labs to create low temperatures.