Q75P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate the molality, molarity, and mole fraction of $\mathit{F}\mathit{e}\mathit{C}{\mathit{l}}_{\mathbf{3}}$  in a 28.8 mass % aqueous solution (d = 1.280 g/mL).

Step-by-Step Solution

Answer

The number of moles of $FeC{l}_3$  and water is 0.18mol and 3.95mol. The molality, the mole fraction, and the molarity of $FeC{l}_3$ is 2.53m, 0.044, and 2.30M respectively.

Concentration of the Solution

A solution can be formed from the dissolution of a solute in a solvent. A solute decides the nature of a solution.

 

A molality can be defined as the ratio of the mass of a solute to the mass of a solution present in kilograms.

 

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

A molarity can be defined as the ratio of the mass of a solute to the volume of a solution present in litres.

 

 $\mathbf{Molarity}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles}}{\mathbf{Volume} \mathbf{of} \mathbf{the} \mathbf{Solution} \mathbf{(} \mathbf{in} \mathbf{Litre} \mathbf{)}}$

A number of moles can be defined as the ratio of the mass of an atom/molecule and the molar mass of the atom/molecule.

 

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

A mole Fraction can be defined as the ratio of the number of moles of a solute to the number of moles of solvent, and the number of a solute.

 

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

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

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

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

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

 

Expression to calculate the Concentration

A mass of 28.8% of ${\mathrm{FeCl}}_3$ means that 28.8g of  ${\mathrm{FeCl}}_3$ I s present in the 100g of solution.

Now, 

 $$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{solution}=\mathrm{Mass} \mathrm{of} \mathrm{solvent},\mathrm{water}+\mathrm{Mass} \mathrm{of} \mathrm{solute} \\ \mathrm{Mass} \mathrm{of} \mathrm{Solvent},\mathrm{water}=\mathrm{Mass} \mathrm{of} \mathrm{solution}--\mathrm{Mass} \mathrm{of} \mathrm{solute},{\mathrm{FeCl}}_3 \\ =100\mathrm{g}-28.8\mathrm{g} \\ =71.2 \mathrm{g}. \end{gathered}$$

 

Mass of Solute, ${\mathrm{Mass}}_{{\mathrm{FeCl}}_3}=28.8\mathrm{g}$

Molar Mass of solute, ${\mathrm{M}}_{{\mathrm{FeCl}}_3}=162\mathrm{g}/\mathrm{mol}$

 $$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{Moles}=\frac{\mathrm{Mass}}{\mathrm{Molar} \mathrm{Mass}} \\ \mathrm{Number} \mathrm{of} \mathrm{Moles} \mathrm{of} {\mathrm{FeCl}}_3=\frac{28.8\mathrm{g}}{162\mathrm{g}/\mathrm{mole}} \\ =0.18 \mathrm{mole}. \end{gathered}$$

Mass of Solvent, ${\mathrm{Mass}}_{\mathrm{Water}}=71.2\mathrm{g}=0.0712\mathrm{kg}$ 

 ${\mathrm{M}}_{\mathrm{water}}=18\mathrm{g}/\mathrm{mol}$

 $$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{Moles}=\frac{\mathrm{Mass}}{\mathrm{Molar} \mathrm{Mass}} \\ \mathrm{Number} \mathrm{of} \mathrm{Moles} \mathrm{of} \mathrm{Water}=\frac{71.2\mathrm{g}}{18\mathrm{g}/\mathrm{mole}} \\ =3.95 \mathrm{mole}. \end{gathered}$$

Number of moles of solute,  ${\mathrm{n}}_{{\mathrm{FeCl}}_3}=0.18\mathrm{mol}$

Number of moles of solvent, ${\mathrm{n}}_{\mathrm{water}}=3.95\mathrm{mol}$ 

 

 $$\begin{gathered} \mathrm{Mole} \mathrm{Fraction}=\frac{\mathrm{Number} \mathrm{of} \mathrm{Moles} \mathrm{of} \mathrm{One} \mathrm{Component}}{\mathrm{Number} \mathrm{of} \mathrm{Moles} \mathrm{of} \mathrm{the} \mathrm{Solvent}+\mathrm{Number} \mathrm{of} \mathrm{Moles} \mathrm{of} \mathrm{Solute}} \\ \mathrm{Mole} \mathrm{Fraction} \mathrm{of} {\mathrm{FeCl}}_3=\frac{0.18}{3.95+0.18} \\ =\frac{0.18}{4.13} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0}44. \end{gathered}$$

$\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{FeCl}}_3=0.18 \mathrm{mol}$

$$\begin{gathered} \mathrm{Molality}=\frac{\mathrm{Number} \mathrm{of} \mathrm{Moles}}{\mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{Solution} ( \mathrm{in} \mathrm{kilogram} )} \\ =\frac{0.18\mathrm{mole}}{0.0712\mathrm{kg}} \\ \mathbf{=}2.53 \mathbf{m}. \end{gathered}$$

$\mathrm{Density} \mathrm{of} \mathrm{solution}=1.28\mathrm{g}/\mathrm{mL}$

$$\begin{gathered} \mathrm{Density} \mathrm{of} \mathrm{Solution}=\frac{\mathrm{Mass}}{\mathrm{Volume}} \\ 1.28\mathrm{g}/\mathrm{mL}=\frac{100\mathrm{g}}{\mathrm{Volume}} \\ \mathrm{Volume}=\frac{100\mathrm{g}}{1.28\mathrm{g}/\mathrm{mL}} \\ \mathbf{=}78.13 \mathbf{mL}. \end{gathered}$$

$$\begin{gathered} Volume of the solution=78.13\mathrm{mL} \\ =0.0.7813L \end{gathered}$$

$$\begin{gathered} \mathrm{Molarity}=\frac{\mathrm{Number} \mathrm{of} \mathrm{Moles}}{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{Solution} ( \mathrm{in} \mathrm{Litre} )} \\ =\frac{0.18\mathrm{mole}}{0.07813\mathrm{L}} \\ =2.30 \mathrm{M}. \end{gathered}$$

The molarity of ${\mathrm{FeCl}}_3=2.30\mathrm{M}.$