1. Mass of ethylene glycol for 500 g of $0.25$ molal solution

Molality is defined as $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$

Let mass of ethylene glycol be $x$ g. Then mass of water (solvent) $= (500 - x)$ g $= \dfrac{500-x}{1000}$ kg.

Moles of ethylene glycol: $\frac{x}{62}$

Given molality $=0.25$: $0.25 = \frac{x/62}{(500-x)/1000}$

$0.25 = \frac{1000x}{62(500-x)}$

$\frac{1}{4} = \frac{1000x}{62(500-x)}$

$62(500-x) = 4000x$

$31000 - 62x = 4000x$

$31000 = 4062x$

$x \approx 7.63\text{ g}$

So, for the molal solution, required mass of ethylene glycol is about $7.63$ g.

2. Mass of ethylene glycol for 250 mL of $0.25$ molar solution

Molarity is defined as $M = \frac{\text{moles of solute}}{\text{volume of solution in L}}$

Given:

• $M = 0.25$
• Volume $=250$ mL $=0.250$ L

Moles of ethylene glycol required: $n = MV = 0.25 \times 0.250 = 0.0625\text{ mol}$

Mass required: $m = n \times \text{molar mass} = 0.0625 \times 62 = 3.875\text{ g}$

So, for the molar solution, required mass of ethylene glycol is $3.875$ g.

3. Required mass ratio

Ratio of masses: $7.63 : 3.875$

$\approx 1.97 : 1$

$\approx 2 : 1$

So the correct option is: $\boxed{\text{C: }2:1}$

4. Comparison with stored correct answer

Stored correct answer is C, which corresponds to $2:1$.

Hence, my derived answer agrees with the stored correct answer.