JEE Chemistry 2023 Question with Solution

Given: Ethylene glycol has molar mass $62 \, \text{g/mol}$. Two aqueous solutions of molarity $0.25$ are to be prepared: one of $500 \, \text{g}$ solution and one of $250 \, \text{mL}$ solution.

Find: The mass ratio of ethylene glycol required in the two cases.

Using the extracted working, assume mass of solvent $\approx$ mass of solution for the first case, so $500 \, \text{g}$ solution is treated as $0.5 \, \text{L}$.

For case I:

$$0.25 = \left(\frac{W_1}{62}\right) \times \left(\frac{1000}{500}\right)$$

Equivalently,

$$\frac{W_1}{62} = 0.25 \times 0.5 = 0.125$$

So,

$$W_1 = 0.125 \times 62 = 7.75 \, \text{g}$$

For case II:

$$0.25 = \left(\frac{W_2}{62}\right) \times \left(\frac{1000}{250}\right)$$

Equivalently,

$$\frac{W_2}{62} = 0.25 \times 0.25 = 0.0625$$

So,

$$W_2 = 0.0625 \times 62 = 3.875 \, \text{g}$$

Therefore,

$$W_1 : W_2 = 7.75 : 3.875 = 2 : 1$$

The correct option is C.

Given: Molarity $M = 0.25 \, \text{mol/L}$, molar mass of ethylene glycol $= 62 \, \text{g/mol}$.

Find: Required mass ratio for the two solutions.

For the $500 \, \text{g}$ solution, the extracted solution treats it approximately as $500 \, \text{mL} = 0.5 \, \text{L}$.

Moles of ethylene glycol:

$$0.25 \times 0.5 = 0.125 \text{ mol}$$

Mass of ethylene glycol:

$$0.125 \times 62 = 7.75 \, \text{g}$$

For the $250 \, \text{mL}$ solution:

Moles of ethylene glycol:

$$0.25 \times 0.25 = 0.0625 \text{ mol}$$

Mass of ethylene glycol:

$$0.0625 \times 62 = 3.875 \, \text{g}$$

Now compare the masses:

$$7.75 : 3.875 = 2 : 1$$

Therefore, the mass ratio of ethylene glycol required is $2:1$, so the correct option is C.

The first solution block on the page labels the option as D, but it simultaneously states the ratio as $2:1$. Since $2:1$ matches option C in the listed options, C is the defensible answer.