QJEEM25Chefa3: JEE Main 2025 Chemistry - Class 11 JEEM25Chefa3 Solution: Density of 3 M NaCl solution is . The molality of the solution…

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

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:

$1074.65 \, \text{g} \div 1000 \approx 1.07465 \, \text{kg}$

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 $2.79 \, \text{m}.$