Molality (m) of 3\ M aqueous solution of NaCl is \ \ \ \ \ . [Given: Density of solution =1.25\ g\ mL -1 , molar mass of NaCl =58.5\ g\ mol -1 ]

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

JEE Main 2024 — 6 April, Shift 2 — Question 61

Molality $(m)$ of $3\ \mathrm{M}$ aqueous solution of $\mathrm{NaCl}$ is $\_\_\_\_\_$ .

[Given: Density of solution $=1.25\ \mathrm{g\ mL^{-1}}$ , molar mass of $\mathrm{NaCl}=58.5\ \mathrm{g\ mol^{-1}}$ ]

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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