Let mass of water initially present = x gm

$$\Rightarrow$$ Mass of sucrose = (1000 $$-$$ x) gm

$$\Rightarrow$$ moles of sucrose = $$\left( {{{1000 - x} \over {342}}} \right)$$

$$ \Rightarrow 0.75 = {{\left( {{{1000 - x} \over {342}}} \right)} \over {\left( {{x \over {1000}}} \right)}} \Rightarrow {x \over {1000}} = {{1000 - x} \over {342 \times 0.75}}$$

$$\Rightarrow$$ 256.5x = 10⁶ $$-$$ 1000x

$$\Rightarrow$$ x = 795.86 gm

$$\Rightarrow$$ moles of sucrose = 0.5969

New mass of H₂O = a kg

$$ \Rightarrow 4 = {{0.5969} \over a} \times 1.86 \Rightarrow $$ a = 0.2775 kg

$$\Rightarrow$$ ice separated = (795.86 $$-$$ 277.5) = 518.3 gm

k_f = 1.86 kg/mol

T$$_f^o$$ = 273 K

solvent : H₂O(625 g)

Solute : 83 g of ethylene glycol

$$\Rightarrow$$ $$\Delta$$T_f = k_f $$\times$$ m

$$\Rightarrow$$ $$\left( {T_f^o - T_f^1} \right) = 1.86 \times {{83/62} \over {625/1000}}$$

$$ \Rightarrow 273 - T_f^1 = {{1.86 \times 83 \times 1000} \over {62 \times 625}} = {{154380} \over {38750}}$$

$$ \Rightarrow 273 - T_f^1 = 4$$

$$ \Rightarrow T_f^1 = 259$$ K

As 0.1 M C₂H₅OH is non-dissociative and rest all salt given are electrolyte so in each case effective molarity > 0.1 so each will have lower freezing point.

Now, $$i = {{final\,moles} \over {initial\,moles}} = {{0.25a + 0.5a + 0.5a} \over {0.5a + 0.5a}}$$

= 1.25 = 125 $$\times$$ 10^$$-$$2