A solution of sodium sulphate contains 92 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 = $23 \ g \ mol^{-1}$.Moles of $Na^{+}$ ions = $\frac{92 \ g}{23 \ g \ mol^{-1}} = 4 \ mol$.Mass of solvent (water) = $1 \ kg$.Molality of $Na^{+}$ ions = $\frac{	ext{moles of } Na^{+}}{	ext{mass of solvent in kg}} = \frac{4 \ mol}{1 \ kg} = 4 \ mol \ kg^{-1}$.

$A$ solution of sodium sulphate contains $92 \ g$ of $Na^{+}$ ions per kilogram of water. The molality of $Na^{+}$ ions in that solution in $mol \ kg^{-1}$ is:

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