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Molality ($$m$$) of $$3M$$ aqueous solution of NaCl is : (Given : Density of solution $$= 1.25 \text{ g mL}^{-1}$$, Molar mass in $$\text{gmol}^{-1}$$: Na $$- 23$$, Cl $$- 35.5$$)
3M NaCl, density 1.25 g/mL. Molar mass NaCl = 58.5 g/mol.
In 1L solution: mass = 1250 g, moles NaCl = 3, mass NaCl = 175.5 g.
Mass of water = 1250 - 175.5 = 1074.5 g = 1.0745 kg.
Molality = 3/1.0745 = 2.79 m.
The correct answer is Option (3): 2.79 m.
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