Question

An aqueous solution containing 12.48g of barium chloride$BaC{l}_2$ in 1000g of water, boils at ${100.0832}^{\circ }C$ .Calculate the degree of dissociation of barium chloride.
$(K_{b}forwater=0.52kgmo{l}^{-1},at.wt.Ba=137,Cl=35.5)$

Answer 1

Hint:
Degree of dissociation is the fraction of original solute molecules that have dissociated. It is usually indicated by Greek symbol $\alpha$.The formula of degree of dissociation is=${\displaystyle \frac{i-1}{n-1}}$

Step by step solution of answer:
- We have been provided with following values-
 Weight of solvent=1000g, here we need to convert it into kg. So, it is equal to 1kg.
 Mass of $BaC{l}_2$added =12.48g
- Now, we need to calculate the molar mass of $BaC{l}_2$=
 $$\begin{array}{rl}& 137\times 2\left(35.5\right)\\ & =208.34g/mol\end{array}$$
$$\mathrm{\Delta }{T}_b=i\times K_b\times {\displaystyle \frac{\text{mass of element}}{\text{molar mass of element}}}$$
  Where, i= vant hoff factor, $K_b$=boiling point constant
   Also,$\mathrm{\Delta }{T}_b=T_f-{T_f}^{\circ }$
 - Where, $T_f$= Final temperature and $T_f^{\circ }$= Initial temperature
  - Firstly, we will calculate van't hoff factor which is equal to,
 $$\begin{array}{rl}& T_f-{T_f}^{\circ }=i\times K_b\times m\\ & (373.0832-373)=i\times 0.52\times {\displaystyle \frac{12.4}{208.34\times 1}}\end{array}$$
The thing here to be noted is that temperature must be converted into Kelvin.
By solving above equation we get the value of, i=2.77
- Van't hoff factor is related to degree of dissociation as:
$$\alpha ={\displaystyle \frac{i-1}{n-1}}$$
$$\begin{array}{rl}& \alpha ={\displaystyle \frac{2.77-1}{3-1}}\\ & \alpha =0.885\end{array}$$
Hence, we can conclude that the degree of dissociation of$BaC{l}_2$ is 0.885.

Additional information:
Degree of dissociation depends on all below factors-
- Nature of electrolyte- weak and strong, and concentration of electrolyte. And usually decreases with increase in concentration.
- As temperature increases, the degree of dissociation increases.
- The nature of solvent polar or non-polar also affects the degree of dissociation.

Note:
- Always remember that while solving this type of question it is mandatory to convert the temperature given in Celsius into Kelvin.
- As the common ion effect decreases then the degree of dissociation increases.
$$\mathrm{\Delta }{T}_b=i\times K_b\times {\displaystyle \frac{\text{mass of element}}{\text{molar mass of element}}}$$
Common ion effect is the difference in equilibrium concentrations between a solution containing a common ion and the same solution in pure water. A common ion is present both in a salt added to a solution and in the solution itself.