Density of 3\ M NaCl solution is 1.25\ g\ mL -1 . The molality of the solution is:

Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2025 — 22 January, Evening Shift — Question 34

Density of $3\ \mathrm{M}$ $\mathrm{NaCl}$ solution is $1.25\ \mathrm{g\ mL^{-1}}$ . The molality of the solution is:

Molality formula, $\mathrm{m = \frac{moles\ of\ solute}{mass\ of\ solvent\ (kg)}}$

Take $1\ L$ solution.

Moles of $\mathrm{NaCl}$ $= \mathrm{3\ mol}$

Mass of solution $= \mathrm{1.25 \times 1000 = 1250\ g}$

Mass of solute $=\mathrm{3 \times 58.5 = 175.5\ g}$

Mass of solvent $= \mathrm{1250 - 175.5 = 1074.5\ g = 1.0745\ kg}$

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

Answer key and solution verified before publishing.

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