JEE Main 2023 — Solutions Question with Solution

The molality of aqueous ethylene glycol solution = 0.25 m

Mass of ethylene glycol required for 1000 g water = $\frac{62}{4} = 15.5 \mathrm{gm}$

Mass of solution = 1015.5 gm

Mass of ethylene glycol in 500 gm solution $= \frac{15.5 \times 500}{1015.5} = 7.63 \mathrm{gm}$

Assuming density of solution as 1 gm/mL.

Mass of ethylene glycol in 250 mL $=\frac{7.63}{2} = 3.815 \mathrm{gm}$

Mass ratio of ethylene glycol for making 500 gm of 0.25 m solution and 250 mL of 0.25 m solution = $2 : 1$

So, the correct option is C