Question
What is boiling point? What is elevation of boiling point? Explain Molal elevation constant and derive It's formula.
✓
Answer
$\rightarrow$ Boiling Point:
$\rightarrow$ "The temperature at which, the vapour pressure of solution is equal to the atmospheric pressure, such temperature is known as boiling point of such solution."
$\rightarrow$ For example, water boils at $373.15 K (100^\circ C)$ because at this temperature the vapour pressure of water is $1.013$ bar $(1$ atmosphere$).$
$\rightarrow$ Vapour pressure of the solvent decreases in the presence of non$-$volatile solute.
$\rightarrow$ The vapour pressure of an aqueous solution of sucrose is less than $1.013$ bar at $373.15 K$. In order to make this solution boil, its vapour pressure must be increased to $1.013$ bar by raising the temperature above the boiling temperature of the pure solvent $($water$).$
$\rightarrow$ Thus, the boiling point of a solution is always higher than that of the boiling point of the pure solvent
$\rightarrow$ Let $T_{ b }^0$ be the boiling point of pure solvent and $T_{ b }$ be the boiling point of solution. The increase in the boiling point $\Delta T_{ b }=T_{ b }-T_{ b }^0$ is known as elevation of boiling point.
$\rightarrow$ Experiments have shown that for dilute solutions the elevation of boiling point $\left(\Delta T_{ b }\right)$ is directly proportional to the molal concentration of the solute in a solution.
$\Delta T_{ b } \propto m$
$\therefore \Delta T_{ b }= K _{ b } . m \ldots .(1)$
Where, $K _{ b }$ is called Boiling Point Elevation Constant or Molal Elevation Constant $($Ebullioscopic Constant$).$
$\rightarrow$ Molal elevation Constant:
$\rightarrow$"Increase in boiling point of a solution prepared by dissolving one gram molar mass of Non volatile solute in one $\ kg$ of solvent is called as Molal elevation constant."
$\rightarrow$ Unit of $K _{ b }= K \cdot \ kg \cdot Mol ^{-1}$
$\rightarrow$ If $w_2$ gram of solute of molar mass $M _2$ is dissolved in $w_1$ gram of solvent, then molality, m of the solution is given by the expression :
$m=\frac{ W _2 \times 1000}{ M _2 \times W _1}$
Substituting value of molality in eq. $(1)$
$\Delta T_{ b }=\frac{K_{ b } \times 1000 \times w_2}{M_2 \times w_1}$
$M_2=\frac{1000 \times w_2 \times K_{ b }}{\Delta T_{ b } \times w_1}$
Where, $w_2=\mathrm{wt}$. of Solute
Where, $w_2 =\text { wt. of Solute }$
$w_1 =\text { wt. of Solvent }$
$M _2 =\text { Molar mass of Solute }$
$\rightarrow$ "The temperature at which, the vapour pressure of solution is equal to the atmospheric pressure, such temperature is known as boiling point of such solution."
$\rightarrow$ For example, water boils at $373.15 K (100^\circ C)$ because at this temperature the vapour pressure of water is $1.013$ bar $(1$ atmosphere$).$
$\rightarrow$ Vapour pressure of the solvent decreases in the presence of non$-$volatile solute.
$\rightarrow$ The vapour pressure of an aqueous solution of sucrose is less than $1.013$ bar at $373.15 K$. In order to make this solution boil, its vapour pressure must be increased to $1.013$ bar by raising the temperature above the boiling temperature of the pure solvent $($water$).$
$\rightarrow$ Thus, the boiling point of a solution is always higher than that of the boiling point of the pure solvent
$\rightarrow$ Let $T_{ b }^0$ be the boiling point of pure solvent and $T_{ b }$ be the boiling point of solution. The increase in the boiling point $\Delta T_{ b }=T_{ b }-T_{ b }^0$ is known as elevation of boiling point.
$\rightarrow$ Experiments have shown that for dilute solutions the elevation of boiling point $\left(\Delta T_{ b }\right)$ is directly proportional to the molal concentration of the solute in a solution.
$\Delta T_{ b } \propto m$
$\therefore \Delta T_{ b }= K _{ b } . m \ldots .(1)$
Where, $K _{ b }$ is called Boiling Point Elevation Constant or Molal Elevation Constant $($Ebullioscopic Constant$).$
$\rightarrow$ Molal elevation Constant:
$\rightarrow$"Increase in boiling point of a solution prepared by dissolving one gram molar mass of Non volatile solute in one $\ kg$ of solvent is called as Molal elevation constant."
$\rightarrow$ Unit of $K _{ b }= K \cdot \ kg \cdot Mol ^{-1}$
$\rightarrow$ If $w_2$ gram of solute of molar mass $M _2$ is dissolved in $w_1$ gram of solvent, then molality, m of the solution is given by the expression :
$m=\frac{ W _2 \times 1000}{ M _2 \times W _1}$
Substituting value of molality in eq. $(1)$
$\Delta T_{ b }=\frac{K_{ b } \times 1000 \times w_2}{M_2 \times w_1}$
$M_2=\frac{1000 \times w_2 \times K_{ b }}{\Delta T_{ b } \times w_1}$
Where, $w_2=\mathrm{wt}$. of Solute
Where, $w_2 =\text { wt. of Solute }$
$w_1 =\text { wt. of Solvent }$
$M _2 =\text { Molar mass of Solute }$
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