JEE Main 2025 — Some Basic Concepts Of Chemistry Question with Solution

$$\text{Molarity of NaCl}=3\text{ M}(\text{3 moles in 1 L of solution})$$

Given the density of the solution is $$1.25\text{g/mL},$$ the total mass of 1 liter (1000 mL) of the solution is:

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

To find the molality, we need the mass of the solvent (water). First, calculate the mass of NaCl in 1 L of solution. The molar mass of NaCl is approximately:

$$23\text{g/mol (Na)}+35.45\text{g/mol (Cl)}\approx 58.45\text{g/mol}$$

Thus, the mass of NaCl is:

$$3\text{mol}\times 58.45\text{g/mol}\approx 175.35\text{g}$$

Now, the mass of the solvent (water) is:

$$1250\text{g (solution)}-175.35\text{g (NaCl)}\approx 1074.65\text{g}$$

Convert the mass of the solvent to kilograms:

$$1074.65\text{g}÷1000\approx 1.07465\text{kg}$$

Molality ($m$) is defined as the number of moles of solute per kilogram of solvent:

$$m=\frac{3\text{mol}}{1.07465\text{kg}}\approx 2.79\text{mol/kg}$$

Thus, the molality of the solution is approximately $$2.79\text{m}.$$