Molality \(( m )\) of \(3 M\) aqueous solution of \(NaCl\) is (Given : Density of solution \(=1.25 ~g mL ^{-1}\), Molar …

Molality \((m)\) is given as

\(m=\frac{{n}_{\text{solute }}}{k{g}_{\text{solvent }}}\)To calculate no. of moles of solute we take the help of molarity and to determine mass of solvent we take the help of density of solution. \(3 M\) solution of \(N a C l\) implies that there are 3 moles of \(N a C l\) in every 1 litre of solution. Considering \(1 L\) of solution, we have

\({n}_{\mathrm{NaCl}}=3\mathrm{mol}\)

Mass of \(1 L\) solution of \(\mathrm{NaCl}=(\) volume \()(\) density )

\(\begin{matrix} & =(1000\mathrm{mL})(1.25{\mathrm{gmol}}^{−1}) \\ & =1250\mathrm{g}\end{matrix}\)

Mass of NaCl in 1L of solution \(\begin{matrix} & =({n}_{\mathrm{NaCl}})({\mathrm{mm}}_{\mathrm{NaCl}})\end{matrix}\)

\(\begin{matrix} & =(3\mathrm{mol})(58.5{\mathrm{gmol}}^{−1}) \\ & =175.5\mathrm{g}\end{matrix}\)

Mass of solvent \(\left({\mathrm{H}}_{2}\mathrm{O}\right)=\) Mass of solution - Mass of solute \(=(1250\mathrm{g})-(175.5\mathrm{g})\)

\(=1074.5g\)

Thus

\(\begin{matrix} & m=\frac{{n}_{\text{solute}}}{{g}_{\text{solv}}}\frac{1000g}{kg} \\ & =\frac{3\mathrm{mol}}{1074.5g}\times \frac{1000g}{\mathrm{kg}} \\ & =2.79molk{g}^{-1}\end{matrix}\\\)

Alternatively \(m=\frac{{M}_{\text{soln }}}{1000d−{M}_{\text{soln }m{m}_{\text{solute }}}}\times 1000\)

\(\begin{matrix} & =\frac{3}{(1000)(1.25)−(3)(58.5)}\times 1000 \\ & =2.79{\mathrm{molkg}}^{−1}\end{matrix}\)