JEE Chemistry 2025 Question with Solution

Given: Molarity of NaCl solution is $3 \, \text{M}$ and density is $1.25 \, \text{g/mL}$.

Find: Molality of the solution.

Take $1 \, \text{L}$ of solution.

From molarity, moles of NaCl in $1 \, \text{L}$ solution are $3$.

Mass of $1 \, \text{L}$ solution:

$$1.25 \, \text{g/mL} \times 1000 \, \text{mL} = 1250 \, \text{g}$$

Mass of NaCl:

$$3 \times 58.44 = 175.32 \, \text{g}$$

Mass of solvent:

$$1250 - 175.32 = 1074.68 \, \text{g} = 1.07468 \, \text{kg}$$

Now molality is:

$$m = \frac{3}{1.07468} = 2.79 \, \text{m}$$

Therefore, the molality is $2.79 \, \text{m}$ and the correct option is A.

The solution also lists option C in one place, but the numerical working clearly gives $2.79 \, \text{m}$.

Given: Molality is defined as moles of solute per kilogram of solvent.

Find: The molality corresponding to a $3 \, \text{M}$ NaCl solution of density $1.25 \, \text{g/mL}$.

For $1 \, \text{L}$ solution, total mass is:

$$1250 \, \text{g}$$

Moles of NaCl present are:

$$3 \, \text{mol}$$

Using molar mass of NaCl as approximately $58.5 \, \text{g/mol}$, mass of solute is:

$$3 \times 58.5 = 175.5 \, \text{g}$$

So mass of solvent is:

$$1250 - 175.5 = 1074.5 \, \text{g} = 1.0745 \, \text{kg}$$

Hence,

$$m = \frac{3}{1.0745} \approx 2.79 \, \text{m}$$

Therefore, the molality is $2.79 \, \text{m}$.