91Ó°ÊÓ

Solution can be formed from the dissolution of the solute in the solvent. The solute decides the nature of the solution.

Molality can be defined as the ratio of the mass of the solute to the mass of the solution present in kilogram measure.

$\mathbf{Molality}\mathbf{=}\frac{\mathbf{Number} \mathbf{of}\mathbf{} \mathbf{Moles}\mathbf{} \mathbf{of}\mathbf{} \mathbf{solute}}{\mathbf{Mass} \mathbf{of}\mathbf{the}\mathbf{} \mathbf{Solution} \mathbf{(} \mathbf{in} \mathbf{kilogram} \mathbf{)}}$

Number of moles can be defined as the ratio of the mass of the atom/molecule and molar mass of the atom/molecule.

$\mathbf{Number}\mathbf{} \mathbf{of}\mathbf{} \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar} \mathbf{Mass}}$

Mole Fraction of solute can be defined as the ratio of the number of moles of the solute to the number of moles of solution.

$\mathbf{Mole} \mathbf{Fraction}\mathbf{=}\frac{\mathbf{Number} \mathbf{of}\mathbf{} \mathbf{Moles}\mathbf{} \mathbf{of} \mathbf{One} \mathbf{Component}}{\mathbf{Number} \mathbf{of} \mathbf{Moles} \mathbf{of}\mathbf{} \mathbf{the} \mathbf{Solvent}\mathbf{+}\mathbf{Number} \mathbf{of}\mathbf{} \mathbf{Moles} \mathbf{of} \mathbf{Solute}}$

Parts by mass: It can be defined as the percentage of the ratio of the mass of solute to the total mass of the solution.

$\mathbf{Mass}\%\mathbf{=}\frac{\mathbf{Mass} \mathbf{of} \mathbf{Solute}}{\mathbf{Mass} \mathbf{of} \mathbf{the}\mathbf{} \mathbf{Solution}}\mathbf{×}\mathbf{100}$

Number of moles of water ${\mathbf{n}}_{\mathbf{water}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{60}\mathbf{mol}$

$$\begin{gathered} \mathbf{Mole} \mathbf{Fraction}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles} \mathbf{of} \mathbf{One} \mathbf{Component}}{\mathbf{Number} \mathbf{of}\mathbf{} \mathbf{Moles} \mathbf{of}\mathbf{} \mathbf{the} \mathbf{Solvent}\mathbf{+}\mathbf{Number} \mathbf{of} \mathbf{Moles} \mathbf{of}\mathbf{} \mathbf{Solute}} \\ \mathbf{Mole}\mathbf{} \mathbf{Fraction} \mathbf{of}\mathbf{} \mathbf{NaCl}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{100} \mathbf{mol}}{\left(\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{+}\mathbf{8}\mathbf{.}\mathbf{6}\right)\mathbf{mol}} \\ \mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{100}}{\mathbf{8}\mathbf{.}\mathbf{7}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0144} \end{gathered}$$

Therefore, mole of fraction of NaCl is 0.0144.

$$\begin{gathered} \mathbf{Number} \mathbf{of} \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar} \mathbf{Mass}} \\ \mathbf{NumberofMoles} \mathbf{of} \mathbf{NaCl}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{58}\mathbf{.}\mathbf{5}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{0}\mathbf{.}\mathbf{100}\mathbf{mol}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{58}\mathbf{.}\mathbf{5}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{Mass} \mathbf{of} \mathbf{NaCl}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{mol}\mathbf{×}\mathbf{58}\mathbf{.}\mathbf{5}\mathbf{g}\mathbf{/}\mathbf{mol} \\ \mathbf{Mass} \mathbf{of} \mathbf{NaCl}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{85}\mathbf{g} \end{gathered}$$

Mass of NaCl, Solute $\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{85}\mathbf{g}$

Number of moles of water $\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{6}\mathbf{mole}$

$$\begin{gathered} \mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar}\mathbf{ }\mathbf{Mass}} \\ \mathbf{NumberofMoles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Water}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{18}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{8}\mathbf{.}\mathbf{6}\mathbf{mol}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{18}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Water}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{6}\mathbf{mol}\mathbf{×}\mathbf{18}\mathbf{g}\mathbf{/}\mathbf{mol} \\ \mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Water}\mathbf{=}\mathbf{154}\mathbf{.}\mathbf{8}\mathbf{g} \end{gathered}$$

Hence, the Mass of water, Solvent $\mathbf{=}\mathbf{154}\mathbf{.}\mathbf{8}\mathbf{g}$

$$\begin{gathered} \mathbf{Mass}\%\mathbf{=}\frac{\mathbf{Mass} \mathbf{of} \mathbf{Solute}}{\mathbf{Mass} \mathbf{of} \mathbf{the} \mathbf{Solution}}\mathbf{×}\mathbf{100} \\ \mathbf{=}\frac{\mathbf{5}\mathbf{.}\mathbf{85}\mathbf{g}}{\mathbf{154}\mathbf{.}\mathbf{8}\mathbf{g}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{85}\mathbf{g}}\mathbf{×}\mathbf{100} \\ \mathbf{=}\frac{\mathbf{5}\mathbf{.}\mathbf{85}\mathbf{g}}{\mathbf{160}\mathbf{.}\mathbf{65}\mathbf{g}}\mathbf{×}\mathbf{100} \\ \mathbf{=}\mathbf{3}\mathbf{.}\mathbf{6}\mathbf{\%} \end{gathered}$$

Mass of Solvent, Water ${\mathbf{Mass}}_{\mathbf{solvent}}\mathbf{=}\mathbf{154}\mathbf{.}\mathbf{8}\mathbf{g}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1548}\mathbf{kg}$

$$\begin{gathered} \mathbf{Molality}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{moles} \mathbf{of} \mathbf{solute}}{\mathbf{Mass} \mathbf{of} \mathbf{the} \mathbf{Solvent} \mathbf{(} \mathbf{in} \mathbf{kg} \mathbf{)}} \\ \mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{mol}}{\mathbf{0}\mathbf{.}\mathbf{1548}\mathbf{kg}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{65}\mathbf{m} \end{gathered}$$

Hence, the Molality is 0.65m