Question

2 molal solution of a weak acid HA has a freezing point of 3.885°C. The degree of dissociation of this acid is ...... x 10^–3. (Round off to the Nearest Integer).

[Given : Molal depression constant of water = 1.85 K kg mol^–1 Freezing point of pure water = 0°C]

Answer 1

ΔT_f = (1 + a) K_f.m

α = 0.05 = 50 × 10^-3

Answer 2

As we know.. ∆Tf = i×kb×m

Using the formula 

3.885 =i × 1.85 × 2

i = 1.05 .......(i)

HA — ^{​​​​​​}H⁺ + A^- ( let dissociation constant = x )

1- x     x     x

i = 1-x+x+x/1.............(ii)

By (i) and (ii) 1.05 = 1 + x 

x = .05

Dissociation constant = .05 × 100 

= 50 Note : the temperature has been given in celcius and depression constant has a unit in Kelvin kg/mol but here we do not convert the temperature to degree Celsius because during the calculation we take difference which is same in both the units     

Answer 3

Given data,

Molality of the weak acid solution is m = 2 molal

Depression in freezing point is \(\triangle T_f=3.885°C\)

Molal depression constant of water is K_f = 1.85 K kg mol^-1

Freezing point of pure water is \(T_F = 0°C\)

Find the van't Hoff factor (i):

\(\triangle T_f = i\times k_f \times m\)

Where, \(\triangle T_f\) is the Depression in freezing point

i is the van't Hoff factor

K_f is the molal depression constant

m is the molality of the solution

Find the degree of dissociation \((\alpha):\)

\(i=1+(n-1)\alpha \)

Where, i is the van't Hoff factor 

\(\alpha\) is the degree of dissociation

n is the number of ions dissociated from the weak acid

\(i.e.,\,HA \rightarrow H^ ++A^ -\)

Here, the number of ions after the dissociation is 2.

Hence, n = 2

By substituting the above-given values in the equation;