Question

An aqueous solution containing \[12.48g\] of barium chloride in $1kg$ of water boils at $373.0832K$ . Calculate the degree of dissociation of barium chloride.
{${K_b}$ for ${H_2}O = 0.52K - gmo{l^{ - 1}}$, Molar mass of $BaC{l_2} = 208.34gmo{l^{ - 1}}$ }
A. $0.165$
B. $0.835$
C. $0.785$
D. None of these

Answer 1

Hint:The degree of dissociation is the fraction of original solute molecules that have dissociated in the solution. More precisely, the degree of dissociation refers to the amount of solute dissociated into ions or radicals per mole in the solution. In the case of very strong acids and bases, the degree of dissociation will be close to $100\% $ .

Complete step by step answer:

The elevation in boiling point is defined as the increase in the boiling point of a liquid when some other miscible liquid is dissolved into it and the boiling point of this solution is greater than the boiling point of pure solvent. It depends upon the degree of dissociation of the solute molecules in the solution and the molality of the solution. Mathematically, it is denoted as:
$\Delta {T_b} = i \times {K_b} \times m$ …..(i)
As per the question, given weight of $BaC{l_2} = 12.48g$
Molar mass of $BaC{l_2} = 208.34gmo{l^{ - 1}}$
Weight of solvent taken = $1kg$
Thus, the molality of a solution is defined as the number of moles of solute present per kilogram of solvent. Mathematically, it is written as:
$m = \dfrac{n}{{{M_{solvent}}(kg)}} = \dfrac{w}{{{M_w}}} \times \dfrac{1}{{{M_{solvent}}(kg)}}$
Substituting the values in the above equation, we have molality as:
$m = \dfrac{{12.48g}}{{208g}} \times \dfrac{1}{1} = 0.06m$
The value of elevation in boiling point is = $\Delta {T_b} = (373.0832 - 373)K = 0.0832K$
Thus, from equation (i), we have:
$i = \dfrac{{\Delta {T_b}}}{{{K_b}m}}$
Substituting the values in the above equation we have:
$i = \dfrac{{0.0832}}{{0.52 \times 0.06}} = 2.67$
We also know that the Van't hoff factor ($i$ ) is equal to the number of moles at equilibrium. Thus, for the dissociation of barium chloride, the reaction can be written as:
$BaC{l_2} \rightleftharpoons B{a^{2 + }} + 2C{l^ - }$
($1 - x$ ) $x$ $2x$
Thus, the total number of moles at equilibrium = $1 - x + x + 2x = 2.67$
Thus, the number of moles of solute dissociated = $x = \dfrac{{2.67 - 1}}{2} = 0.835$
Hence, the degree of dissociation = $0.835 \times 100 = 83.5\% $
The correct option is B. $0.835$ .

Note:
The Van't Hoff factor is the ratio of the actual concentration of particles produced when the substance is dissolved in the solution and the concentration of a substance as calculated with the help of its mass. For most of the non-electrolytes that get dissolved in water, the Van't Hoff factor is equal to one.