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Question
If the density of some lake water is 1.25 g mL−1 and contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake.
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Solution
Given:
Mass of Na+ ions = 92 g
Mass of water (solvent) = 1 kg
Molar mass of Na+ = 23 g/mol
Formula:
Molality (m) = `"Moles of solute"/"Mass of solvent (kg)"`
Solution:
Number of moles present in 92 g of Na+ ions = `"92 g"/(23 "g mol"^(-1))` ...(as atomic mass of Na = 23)
= 4 mole
Molality (m) = `"Moles of solute"/"Mass of solvent (kg)"`
Therefore, molality of Na+ ions in the lake = `(4 "mole")/ (1 "kg")`
= 4 mol/kg
Hence, molality of lake water = 4 m
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