JEE Main 2025 Jan 22 Shift 2, Chemistry Q11: Types Concentration And Solubility of Solutions

JEE Main2025Jan 22, Shift 2Chemistry

Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is :

Solution

Take 1 L of solution as a convenient basis.

Mass of solution $= 1000 \times 1.25 = 1250 g$ .

Moles of NaCl in 1 L of a 3 M solution $= 3 m o l$ , so mass of NaCl $= 3 \times 58.5 = 175.5 g$ .

Mass of water (solvent) $= 1250 - 175.5 = 1074.5 g = 1.0745 k g$ .

Molality $= \frac{moles of solute}{kg of solvent} = \frac{3}{1.0745} \approx 2.79 m$ .

Concept behind this question

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