Name: Non-ideal Solutions
Uploaded: 2019-07-19
Duration: 8 min 43 s
Description: We know that an ideal solution is a theoretical concept. So, what are non-ideal solutions? Why do they show a deviation from Raoult's law? Let's explore!

Question

For a non-volatile solute's solution, prove that

∆ T_{b} = K_{b} . m

.

Accepted Answer

Elevation in boiling point, $∆ T_{b} = T_{b} - T_{o}$
Here, $T_{b}$ is the boiling point of the solution and $T_{o}$ is the boiling point of solvent.
The elevation in boiling point is directly proportional to lowering in vapour pressure.
$∆ T_{b} \propto ∆ P$

According to Raoult's law, lowering in vapour pressure is directly proportional to the mole fraction of solute, hence,
$∆ T_{B} \propto χ_{B}$ .

$∆ T_{B} = K χ_{B}$ .

But, $χ_{B} = \frac{\frac{W_{B}}{M_{B}}}{\frac{W_{A}}{M_{A}} + \frac{W_{B}}{M_{B}}}$

Hence, $∆ T_{B} = K \frac{\frac{W_{B}}{M_{B}}}{\frac{W_{A}}{M_{A}} + \frac{W_{B}}{M_{B}}}$

For dilute solutions, $\frac{W_{B}}{M_{B}} < < < \frac{W_{A}}{M_{A}}$

$\Rightarrow ∆ T_{B} = K \frac{\frac{W_{B}}{M_{B}}}{\frac{W_{A}}{M_{A}}} \Rightarrow {KM}_{A} \frac{W_{B}}{M_{B} \times W_{A}}$

$\Rightarrow ∆ T_{b} = K_{b} m$ ( $∵ {KM}_{A} = K_{b}$ )

Where $m$ is molality of solution and $K_{b}$ is molal elevation constant.

For a non-volatile solute's solution, prove that $∆ T_{b} = K_{b} . m$ .

Important Questions on Solutions

Learn and Prepare for any exam you want !

For a non-volatile solute's solution, prove that ∆Tb=Kb.m.

Important Questions on Solutions

Learn and Prepare for any exam you want !

For a non-volatile solute's solution, prove that $∆ T_{b} = K_{b} . m$ .