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

Addition of solute decreases the vapour pressure as some sites of the surface are occupied by solute particles resulting in decreased surface area. However, addition of solvent, i.e., dilution increases the surface area of the liquid surface, thus results in increased vapour pressure. Hence, addition of water to the aqueous solution of (1 molal) KI results in increased vapour pressure.

Depression in freezing point,

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

m = $${{{w_B}} \over {{M_B}}} \times {{1000} \over {{W_A}}}$$

= $${{68.5 \times 1000} \over {342 \times 1000}}$$

$$\Delta $$T_f = 1.86 $$ \times $$ $${{68.5} \over {342}}$$

= 0.372 ^oC

$$ \therefore $$ T_f = 0 - 0.372 ^oC = - 0.372 ^oC

The number of moles of ions produced by 1 mol of ionic compound = i

Given, m = 0.0020 m

$$\Delta $$T_f = 0^oC – 0.00732^oC = – 0.00732^oC

K_f = – 1.86 ^oC

$$\Delta $$T_f = ik_fm

$$ \Rightarrow $$ i = $${{0.00732} \over {1.86 \times 0.0020}}$$ = 2