Question

An aqueous solution containing 12.48g of barium chloride in 1.0 kg of water boils at 373.0832 K. calculate the degree of dissociation of barium chloride. (given, Kb for H2O=0.52 Kmolâ -1, molar mass of BaCl2=208.34 g mol-1)

Answer 1

Answer: Degree of dissociation is 0.0885

Explanation:-

Weight of solvent =  1.0 kg

Molar mass of $BaCl_2$= 208.34 g/mol

Mass of $BaCl_2$ added = 12.48 g

$\Delta T_b=i\times K_b\times \frac{\text{mass of barium chloride}}{\text{molar mass of barium chloride}\times \text{weight of solvent in kg}}$

$\Delta T_b=T_f-T^{0}_f$ = elevation in boiling point

i= vant hoff factor

$K_b$ = boiling point constant

$(373.0832-373)K=i\times 0.52\times \frac{12.4}{208.34\times 1.0}$

$0.0832$=i\times 0.03[/tex]

 $i=2.77$    

Vant hoff's factor(i) is related to degree of dissociation $(\alpha)$ as:

 $\alpha =\frac{i-1}{n-1}=\frac{2.77-1}{3-1}$

$\alpha=0.885$  

Answer 2

Answer: The degree of dissociation of barium chloride is 84 %.

Explanation:

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}}{M_{solute}\times W_{solvent}\text{ (in kg)}}$

where,

Boiling point of water = 373 K

Boiling point of barium chloride solution = 373.0832 K  

i = Vant hoff factor = ?

$K_b$ = molal boiling point elevation constant = 0.52 K/m

$m_{solute}$ = Given mass of solute (barium chloride) = 12.48 g

$M_{solute}$ = Molar mass of solute (barium chloride) = 208.34  g/mol

$W_{solvent}$ = Mass of solvent (water) = 1.0 kg

Putting values in above equation, we get:

$(373.0832-373)K=i\times 0.52K/m\times \frac{12.48g}{208.34g/mol\times 1.0kg}\\\\i=2.68$

The relationship between Van't Hoff factor and degree of dissociation is given as:

$\alpha=\frac{i-1}{n-1}\times 100$

where,

n = number of ions = 3

i = Van't Hoff factor = 2.68

Putting values in above equation, we get:

$\alpha=\frac{2.68-1}{3-1}\times 100\\\\\alpha=84\%$

Hence, the degree of dissociation of barium chloride is 84 %.