If the density of some lake water is 1.25, g, | vedclass

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(B) The molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.Given mass of $Na^{+}$ ions $= 92\, g$.Molar mass of $Na^{+

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$4$

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(B) The molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.Given mass of $Na^{+}$ ions $= 92\, g$.Molar mass of $Na^{+}$ ions $= 23\, g\, mol^{-1}$.Number of moles of $Na^{+}$ ions $= \frac{92\, g}{23\, g\, mol^{-1}} = 4\, mol$.Mass of solvent (water) $= 1\, kg$.Molality $(m)$ $= \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{4\, mol}{1\, kg} = 4\, m$.

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. (in $, m$)

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