Question
Derive the relation between molar mass of the solute and boiling point elevation.
✓
Answer
The boiling point elevation, $\Delta T_b$ of a solution is directly proportional to molality (m) of the solution.
$\begin{aligned}
& \therefore \Delta T_b \propto m \\
& \Delta T_b=K_b m
\end{aligned}$
where $K _b$ is a proportionality constant
If $m=1$ molal, then
$\Delta T_b=K_b$
where $K_b$ is called molal elevation constant.
The molality of the solution is given by,
Number of moles of the solute, $m=\frac{\text { Number of moles of the solute }}{\text { Weight of the solvent in } kg }$
Let $W _1=$ Weight (in gram) of a solvent,
$W _2=$ Weight (in gram) of a solute
$M _2=$ Molecular weight of the solute
Then the molality ( $m$ ) of the solution is given by
If the weights and molecular weight are expressed in $kg$, then,
$\Delta T_{ b }=K_{ b } \times \frac{W_2}{W_1 M_2}\
$\begin{aligned}
& \therefore \Delta T_b \propto m \\
& \Delta T_b=K_b m
\end{aligned}$
where $K _b$ is a proportionality constant
If $m=1$ molal, then
$\Delta T_b=K_b$
where $K_b$ is called molal elevation constant.
The molality of the solution is given by,
Number of moles of the solute, $m=\frac{\text { Number of moles of the solute }}{\text { Weight of the solvent in } kg }$
Let $W _1=$ Weight (in gram) of a solvent,
$W _2=$ Weight (in gram) of a solute
$M _2=$ Molecular weight of the solute
Then the molality ( $m$ ) of the solution is given by
If the weights and molecular weight are expressed in $kg$, then,
$\Delta T_{ b }=K_{ b } \times \frac{W_2}{W_1 M_2}\
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