Question

$1\mathrm{molal}$ aqueous solution of an electrolyte ${\mathrm{A}}_2{\mathrm{B}}_3$ is $60\%$ ionised. The boiling point of the solution at $1\mathrm{atm}$ is $\mathrm{\_\_\_\_\_}\mathrm{K}$.

Answer 1

Step 1: Given data:

Molality of the solution is $\mathrm{m}=1\mathrm{molal}$

Type of electrolyte is ${\mathrm{A}}_2{\mathrm{B}}_3$

Percentage of ionisation is $\mathrm{\alpha }=60\%$

$\Rightarrow \mathrm{\alpha }=0.6$

Pressure is $\mathrm{P}=1\mathrm{atm}$

Step 2: Write the balanced equation of the dissociation of the ${\mathbf{A}}_{\mathbf{2}}{\mathbf{B}}_{\mathbf{3}}$ electrolyte:

${\mathrm{A}}_2{\mathrm{B}}_3\to 2{\mathrm{A}}^{+3}+3{\mathrm{B}}^{-2}$

Number of ${\mathrm{A}}^{+3}$ ions is $2$

Number of $B^{-2}$ ions is $3$

Total number of ions is $\mathrm{n}=2+3$

$\Rightarrow \mathrm{n}=5$

Step 3: Find the value of van't Hoff factor $\left(i\right)$:

$$\begin{gathered} \mathrm{i}=1+\left(\mathrm{n}-1\right)\times \mathrm{\alpha } \\ \Rightarrow \mathrm{i}=1+\left(5-1\right)\times 0.6 \\ \Rightarrow \mathrm{i}=1+2.4 \\ \Rightarrow \mathrm{i}=3.4 \end{gathered}$$

Step 4: Find the value of elevation in boiling point $\left(∆T_b\right)$ of the given solution:

$∆{\mathrm{T}}_{\mathrm{b}}={\mathrm{k}}_{\mathrm{b}}\times \mathrm{m}\times \mathrm{i}$

Where, $∆{\mathrm{T}}_{\mathrm{b}}$ is the elevation in boiling point

${\mathrm{k}}_{\mathrm{b}}$ is the molal boiling point constant$=0.52\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{-1}$

$\mathrm{m}$ is the molality of the solution $=1\mathrm{molal}$

$\mathrm{i}$ is the van't Hoff factor $=3.4$

Substitute all the values in the above equation;

$$\begin{gathered} \Rightarrow ∆{\mathrm{T}}_{\mathrm{b}}=0.52°\mathrm{C}{\mathrm{mol}}^{-1}\times 1\mathrm{molal}\times 3.4 \\ \Rightarrow ∆{\mathrm{T}}_{\mathrm{b}}=1.768°\mathrm{C} \end{gathered}$$

Step 5: Find the boiling point of $\mathbf{1}\mathbf{}\mathbf{molal}$ aqueous solution $\left[{\left(T_b\right)}_{\mathrm{solution}}\right]$:

$∆{\mathrm{T}}_{\mathrm{b}}={\left({\mathrm{T}}_{\mathrm{b}}\right)}_{\mathrm{solution}}-{\left[{\left({\mathrm{T}}_{\mathrm{b}}\right)}_{{\mathrm{H}}_2\mathrm{O}}\right]}_{\mathrm{solution}}$

Where, $∆{\mathrm{T}}_{\mathrm{b}}$ is the elevation in boiling point

${\left({\mathrm{T}}_{\mathrm{b}}\right)}_{\mathrm{solution}}$ is the boiling point of the solution

${\left[{\left({\mathrm{T}}_{\mathrm{b}}\right)}_{{\mathrm{H}}_2\mathrm{O}}\right]}_{\mathrm{solution}}$ is the boiling point of the water in the solution

$$\begin{gathered} \Rightarrow 1.768°\mathrm{C}={\left({\mathrm{T}}_{\mathrm{b}}\right)}_{\mathrm{solution}}-100°\mathrm{C} \\ \Rightarrow {\left({\mathrm{T}}_{\mathrm{b}}\right)}_{\mathrm{solution}}=1.768+100 \\ \Rightarrow {\left({\mathrm{T}}_{\mathrm{b}}\right)}_{\mathrm{solution}}=101.768°\mathrm{C} \\ \Rightarrow {\left({\mathrm{T}}_{\mathrm{b}}\right)}_{\mathrm{solution}}=101.768+273\mathrm{K} \\ \Rightarrow {\left({\mathrm{T}}_{\mathrm{b}}\right)}_{\mathrm{solution}}=374.768\mathrm{K} \end{gathered}$$

Hence, the boiling point of $\mathbf{1}\mathbf{}\mathbf{molal}$ aqueous solution is ${\left(T_b\right)}_{\mathbf{solution}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{374}\mathbf{.}\mathbf{768}\mathbf{}\mathbf{K}$.