Question

What is the molality of a solution consisting of 44.0 mL of benzene ( ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{6}}$; d = 0.877 g/mL) in 167 mL of hexane (data-custom-editor="chemistry" ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{14}}$ ; d = 0.660 g/mL)?

Short answer

The molality of the solution is 4.4 m.

Step-by-step solution

Concentration of the Solution

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.

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

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

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

Density can be defined as the ratio of the mass of the matter to the volume of the matter.

$\mathbf{Density}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Volume}}$

Expression to calculate the Concentration

a) Volume of benzene$\mathbf{=}\mathbf{44}\mathbf{mL}$

Density of the benzene $\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{877}\mathbf{g}\mathbf{/}\mathbf{mL}$

$$\begin{gathered} \mathbf{Density} \mathbf{of} \mathbf{Benzene}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Volume}} \\ \mathbf{0}\mathbf{.}\mathbf{877} \mathbf{g}\mathbf{/}\mathbf{mL}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{44}\mathbf{mL}} \\ \mathbf{Mass}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{877} \mathbf{g}\mathbf{/}\mathbf{mL}\mathbf{\times }\mathbf{44} \mathbf{mL} \\ \mathbf{Mass}\mathbf{=}\mathbf{38}\mathbf{.}\mathbf{6}\mathbf{g} \end{gathered}$$

Mass of Benzene$\mathbf{=}\mathbf{38}\mathbf{.}\mathbf{6}\mathbf{g}$