Question

A solution of $10$g of a non-volatile binary electrolyte (mat.wt.=$100$) in $500$g of water freezes at $- {0.74^o}C$. What is the degree of ionisation?
[${k_f}$ of water=$1.85$K$molalit{y^{ - 1}}$]
a. $50\%$
b. $75\%$
c. $100\%$
d. $0\%$

Answer 1

Depression in freezing point is a colligative property. Colligative properties are directly proportional to the number of particles of solute.

Complete step by step answer:
When a solute is added to a solvent, the freezing point of solvent decreases. This is called depression in freezing point. Depression in freezing point is directly proportional to the number of particles of the solute in solution.
We can write, $\vartriangle {T_f} = i{k_f}m$
Where,
$\vartriangle {T_f}$= depression in freezing point
$i$ = vant-Hoff factor
${k_f}$= molal depression constant
m = molality of solution
Water usually freezes at $273K$. The given solution freezes at $- {0.74^o}C$.
Freezing point of solution = $- {0.74^o}C$ = $( - 0.74 + 273)K = 272.26K$
Hence depression in freezing point is given by,
$\vartriangle {T_f}$= $273 - 272.26 = 0.74K$
Given ${k_f}$ of water=$1.85$K$molalit{y^{ - 1}}$
Molality can be calculated by,
Molality (m) = $\dfrac{{{w_B} \times 1000}}{{{M_B} \times {W_A}}}$
Where,
${w_B}$ = weight of solute in grams
${M_B}$ = molecular weight of solute
${W_A}$ = weight of solvent in grams
Given,
${w_B}$ = $10$g
${M_B}$ = $100$ g
${W_A}$ = $500$g
Let us substitute these values into the equation of molality.
Molality (m) = $\dfrac{{10 \times 1000}}{{100 \times 500}} = 0.2$m
Now let us find out the value of i using the equation,
$\vartriangle {T_f} = i{k_f}m$
$i = \dfrac{{\vartriangle {T_f}}}{{{k_f}m}} = \dfrac{{0.74}}{{1.85 \times 0.2}} = 2$
We got the value of $i$ as 2. We need to find the degree of ionisation ($\alpha$). The relation between $i$ and $\alpha$is,
$\alpha = \dfrac{{i - 1}}{{n - 1}}$
Where n is the number of particles in which one molecule of electrolyte dissociate. For binary electrolytes,
n = $2$
Let us substitute the value of $i$ and n to the above equation to find out $\alpha$.
$\alpha = \dfrac{{2 - 1}}{{2 - 1}} = \dfrac{1}{1} = 1$
The degree of ionisation ($\alpha$) is 1. Converting it into percentage, we get the degree of ionisation as $100\%$.
Hence the correct option is C.

Note:
-Things that should be remembered while solving this type of questions are:
-The solute should be non-volatile.
-We should be aware of the nature of the solute. i.e. we should know if the solute undergo association or dissociation in the given solvent.