We know that $$\Delta $$T_f = i × K_f × m

Here i is van’t Hoff’s factor.

i for weak acid is 1 + $$\alpha $$.

Here $$\alpha $$ is degree of dissociation i.e., 30/100 = 0.3

$$ \therefore $$ i = 1 + $$\alpha $$ = 1 + 0.3 = 1.3

$$ \therefore $$ T_f = i × K_f × m = 1.3 × 1.86 × 0.1 = 0.24

$$ \therefore $$ Freezing point = – 0.24

According to depression in freezing point,

$$\Delta $$T_f = i $$ \times $$ K_f $$ \times $$ m

= i $$ \times $$ K_f $$ \times $$ $${{{w_B} \times 1000} \over {{m_B} \times {w_A}}}$$

Given : $$\Delta $$T_f = 3.82, K_f = 1.86,

w_B = 5, m_B = 142, w_A = 45

i = $${{\Delta T \times {m_B} \times {w_A}} \over {{K_f} \times {w_B} \times 1000}}$$

= $${{3.82 \times 142 \times 45} \over {1.86 \times 5 \times 1000}}$$

= 2.63

From the value of van’t Hoff factor i it is possible to determine the degree of dissociation or association. In case of dissociation, i is greater than 1 and in case of association i is less than 1.

Osmotic pressure, $$\pi $$ = CRT

$$ \Rightarrow $$ $$\pi $$ = $${n \over V}$$RT

$$ \Rightarrow $$ $$\pi $$V = $${w \over M}$$RT

M = $${{wRT} \over {\pi V}}$$

= $${{1.26 \times 0.083 \times 300} \over {2.57 \times {{10}^{ - 3}} \times {{200} \over {1000}}}}$$

= $${{1.26 \times 0.083 \times 300} \over {2.57 \times {{10}^{ - 3}} \times 0.2}}$$

= 61038 g mol^–1