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.
$\left( {{K}_{b}}forwater=0.52kgmo{{l}^{-1}},at.wt.Ba=137,Cl=35.5 \right)$

Answer 1

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=$\dfrac{i-1}{n-1}$

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 $BaC{{l}_{2}}$=

& 137\times 2\left( 35.5 \right) \\\ & =208.34g/mol \\\ \end{aligned}$$ $$\Delta {{T}_{b}}=i\times {{K}_{b}}\times \dfrac{\text{mass of element}}{\text{molar mass of element}}$$ Where, i= vant hoff factor, ${{K}_{b}}$=boiling point constant Also,$\Delta {{T}_{b}}={{T}_{f}}-{{T}_{f}}^{\circ }$ \- Where, ${{T}_{f}}$= Final temperature and $T_{f}^{\circ }$= Initial temperature \- Firstly, we will calculate van't hoff factor which is equal to, $$\begin{aligned} & {{T}_{f}}-{{T}_{f}}^{\circ }=i\times {{K}_{b}}\times m \\\ & \left( 373.0832-373 \right)=i\times 0.52\times \dfrac{12.4}{208.34\times 1} \\\ \end{aligned}$$ The thing here to be noted is that temperature must be converted into Kelvin. By solving above equation we get the value of, i=2.77 \- Van't hoff factor is related to degree of dissociation as: $$\alpha =\dfrac{i-1}{n-1}$$ $$\begin{aligned} & \alpha =\dfrac{2.77-1}{3-1} \\\ & \alpha =0.885 \\\ \end{aligned}$$ Hence, we can conclude that the degree of dissociation of$BaC{{l}_{2}}$ is 0.885. **Additional information:** Degree of dissociation depends on all below factors- \- Nature of electrolyte- weak and strong, and concentration of electrolyte. And usually decreases with increase in concentration. \- As temperature increases, the degree of dissociation increases. \- The nature of solvent polar or non-polar also affects the degree of dissociation. **Note:** \- Always remember that while solving this type of question it is mandatory to convert the temperature given in Celsius into Kelvin. \- As the common ion effect decreases then the degree of dissociation increases. $$\Delta {{T}_{b}}=i\times {{K}_{b}}\times \dfrac{\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.