
What is the boiling point of sugar solution at atmospheric pressure ? How do we calculate boiling point elevation of sugar solution?
Answer
550.5k+ views
Hint: Concentration of non volatile solute plays a major role in determining the boiling point. As more solute is added, there is very less space for the solvent to escape and this increases the boiling point.
Complete answer:
In order to answer our question, we need to learn about the properties of elevation in boiling point. We know that dissolution of non volatile solute in a liquid causes lowering of its vapour pressure. The boiling point of a liquid may be defined as the temperature at which its vapour pressure becomes equal to atmospheric pressure. At temperature ${{T}_{0}}$, the vapour pressure of the pure solvent becomes equal to the atmospheric pressure. Therefore, ${{T}_{0}}$ is the boiling point of the pure solvent. At each temperature the vapour pressure of solution is lower than that of the pure solvent and thus, the vapour pressure curve for the solution runs below that of the pure solvent. To find the difference in boiling point, we have the formula:
\[\Delta {{T}_{b}}={{K}_{b}}\times m\]
Where ${{K}_{b}}$ is called boiling point-elevation constant or molal elevation constant or molal ebullioscopic constant of the solvent, and m is molality.
Now, to find the boiling point , let us assume that 2000 grams of sugar is dissolved in 1000 grams water. Sucrose ${{C}_{6}}{{H}_{12}}{{O}_{6}}$ has a molar mass of $342g\,mo{{l}^{-1}}$. Now, the molality of sucrose in the solution is:
\[\begin{align}
& m=\dfrac{given\,mass}{molar\,mass}\times \dfrac{1000}{mass\,of\,solvent} \\
& \Rightarrow m=\dfrac{2000}{343.3}\times \dfrac{1000}{1000} \\
& \Rightarrow m=5.48\,mol\,k{{g}^{-1}} \\
\end{align}\]
Also, ${{K}_{b}}$ is ${{0.515}^{0}}C\,mol\,k{{g}^{-1}}$.
$\Delta {{T}_{b}}=0.515\times 5.483$
=${{2.824}^{0}}C$
So, the final boiling point obtained is ${{(100+2.824)}^{0}}C={{102.824}^{0}}C$. This is how we calculate the boiling point of sucrose.
Note:
It is to be noted that the actual formula for finding the difference in boiling point is $\Delta {{T}_{b}}=i\times {{K}_{b}}\times m$, where ‘i’ is the Van't Hoff factor and is only applicable for electrolytic solutions. In our case, sugar is non electrolytic o, Van't Hoff factor is considered 1.
Complete answer:
In order to answer our question, we need to learn about the properties of elevation in boiling point. We know that dissolution of non volatile solute in a liquid causes lowering of its vapour pressure. The boiling point of a liquid may be defined as the temperature at which its vapour pressure becomes equal to atmospheric pressure. At temperature ${{T}_{0}}$, the vapour pressure of the pure solvent becomes equal to the atmospheric pressure. Therefore, ${{T}_{0}}$ is the boiling point of the pure solvent. At each temperature the vapour pressure of solution is lower than that of the pure solvent and thus, the vapour pressure curve for the solution runs below that of the pure solvent. To find the difference in boiling point, we have the formula:
\[\Delta {{T}_{b}}={{K}_{b}}\times m\]
Where ${{K}_{b}}$ is called boiling point-elevation constant or molal elevation constant or molal ebullioscopic constant of the solvent, and m is molality.
Now, to find the boiling point , let us assume that 2000 grams of sugar is dissolved in 1000 grams water. Sucrose ${{C}_{6}}{{H}_{12}}{{O}_{6}}$ has a molar mass of $342g\,mo{{l}^{-1}}$. Now, the molality of sucrose in the solution is:
\[\begin{align}
& m=\dfrac{given\,mass}{molar\,mass}\times \dfrac{1000}{mass\,of\,solvent} \\
& \Rightarrow m=\dfrac{2000}{343.3}\times \dfrac{1000}{1000} \\
& \Rightarrow m=5.48\,mol\,k{{g}^{-1}} \\
\end{align}\]
Also, ${{K}_{b}}$ is ${{0.515}^{0}}C\,mol\,k{{g}^{-1}}$.
$\Delta {{T}_{b}}=0.515\times 5.483$
=${{2.824}^{0}}C$
So, the final boiling point obtained is ${{(100+2.824)}^{0}}C={{102.824}^{0}}C$. This is how we calculate the boiling point of sucrose.
Note:
It is to be noted that the actual formula for finding the difference in boiling point is $\Delta {{T}_{b}}=i\times {{K}_{b}}\times m$, where ‘i’ is the Van't Hoff factor and is only applicable for electrolytic solutions. In our case, sugar is non electrolytic o, Van't Hoff factor is considered 1.
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