Q77P - Chemistry: Molecular Nature Of Matter And Change

Question

An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/mL; ɱ = 62.07 g/mole) and water (d = 1.00 g/mL) at ${\mathbf{20}}^{\mathbf{o}}\mathbf{C}$. The density of the mixture is 1.070g/mL. Express the concentration of ethylene glycol as: 

(a) Volume percent                        

(b) Mass percent 

(c) Molarity                                   

 (d) Molality 

(e) Mole fraction                                     

 

Step-by-Step Solution

Answer

a. The volume percentage of Ethylene glycol is 50% because 50 mL of ethylene glycol (the solvent) is present in 100 mL of the solution. 

b. The mass percentage of the ethylene glycol is 52.1%.

c. The molarity of the ethylene glycol is 8.974 M.

d. The molality of the ethylene glycol is 17.95 m.

e. The mass percentage of the ethylene glycol is 0.245%.

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 to 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 a 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.

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

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

 

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

 

A 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. 

Volume percentage: 50% of ethylene glycol means that 50 mL of ethylene glycol is present in 100 mL of the mixture.

 

Density of the Ethylene glycol $\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{114}\mathbf{g}\mathbf{/}\mathbf{mL}$

Volume of the Ethylene glycol $\mathbf{=}\mathbf{50}\mathbf{mL}$

$$\begin{gathered} \mathbf{Density} \mathbf{of} \mathbf{Ethylene} \mathbf{Glycol}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Volume}} \\ \mathbf{1}\mathbf{.}\mathbf{114}\mathbf{g}\mathbf{/}\mathbf{mL}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{50}\mathbf{mL}} \\ \mathbf{Mass}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{114}\mathbf{g}\mathbf{/}\mathbf{mL}\mathbf{\times }\mathbf{50}\mathbf{mL} \\ \mathbf{Mass} \mathbf{of} \mathbf{Ethylene} \mathbf{Glycol}\mathbf{=}\mathbf{55}\mathbf{.}\mathbf{7} \mathbf{g}. \end{gathered}$$

Density of the solution $\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{07}\mathbf{g}\mathbf{/}\mathbf{mL}$

Volume of the solution $\mathbf{=}\mathbf{100}\mathbf{mL}$

$$\begin{gathered} \mathbf{Density} \mathbf{of} \mathbf{Solution}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Volume}} \\ \mathbf{1}\mathbf{.}\mathbf{07}\mathbf{g}\mathbf{/}\mathbf{mL}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{100}\mathbf{mL}} \\ \mathbf{Mass}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{07}\mathbf{g}\mathbf{/}\mathbf{mL}\mathbf{\times }\mathbf{100}\mathbf{mL} \\ \mathbf{=}107 \mathbf{g}. \end{gathered}$$

 

b. 

Mass Percentage of Ethylene Glycol:

$$\begin{gathered} \mathbf{mass}\mathbf{\%}\mathbf{=}\frac{\mathbf{Mass} \mathbf{of} \mathbf{Solute}}{\mathbf{Mass} \mathbf{of} \mathbf{the} \mathbf{Solution}}\mathbf{\times }\mathbf{100} \\ \mathbf{=}\frac{\mathbf{55}\mathbf{.}\mathbf{7}}{\mathbf{107}\mathbf{g}}\mathbf{\times }\mathbf{100} \\ \mathbf{=}\mathbf{52}\mathbf{.}\mathbf{1}\mathbf{\%} \end{gathered}$$ 

 

c. 

The mass of Ethylene Glycol  $\mathbf{=}\mathbf{55}\mathbf{.}\mathbf{7}\mathbf{g}$

The molar Mass of Ethylene Glycol $\mathbf{=}\mathbf{62}\mathbf{g}\mathbf{/}\mathbf{mole}$

$$\begin{gathered} \mathbf{Number} \mathbf{of} \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar} \mathbf{Mass}} \\ \mathbf{=}\frac{55.7\mathbf{g}}{\mathbf{62}\mathbf{g}\mathbf{/}\mathbf{mole}} \\ \mathbf{=}\mathbf{0}.8974\mathbf{mole}. \end{gathered}$$

Volume of the solution $\mathbf{=}\mathbf{100}\mathbf{mL}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{L}$

The molarity of Ethylene Glycol:

$$\begin{gathered} \mathbf{Density} \mathbf{of} the s\mathbf{olution}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Volume}} \\ \mathbf{1}\mathbf{.}\mathbf{0}\mathbf{g}\mathbf{/}\mathbf{mL}\mathbf{=}\frac{\mathbf{Mass}}{5\mathbf{0}\mathbf{mL}} \\ \mathbf{Mass}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{g}\mathbf{/}\mathbf{mL}\mathbf{\times }5\mathbf{0}\mathbf{mL} \\ \mathbf{=}50\mathbf{g}. \end{gathered}$$

Mass of the solvent, water $\mathbf{=}\mathbf{50}\mathbf{g}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{kg}$

$$\begin{gathered} \mathbf{Molality}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles}}{\mathbf{Mass} \mathbf{of} \mathbf{the} \mathbf{Solution} \mathbf{(} \mathbf{in} \mathbf{kilogram} \mathbf{)}} \\ \mathbf{=}\frac{\mathbf{0}.8974 \mathbf{mole}}{\mathbf{0}\mathbf{.}\mathbf{0}5 \mathbf{kg}} \\ \mathbf{Mol}\mathbf{=}17.95 \mathbf{m}. \end{gathered}$$

e.

Number of moles of Ethylene Glycol  $\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8974}\mathbf{mole}$

The mass of the solvent, water $\mathbf{=}\mathbf{50}\mathbf{g}$ 

The molar mass of the solvent, water  $\mathbf{=}\mathbf{18}\mathbf{g}\mathbf{/}\mathbf{mole}$

$$\begin{gathered} \mathbf{Number} \mathbf{of} \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar} \mathbf{Mass}} \\ \mathbf{=}\frac{\mathbf{50}\mathbf{g}}{\mathbf{18}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{776} \mathbf{mol}. \end{gathered}$$

Number of moles of the solvent, water $\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{776}\mathbf{mole}$

$$\begin{gathered} \mathbf{Mole} \mathbf{Fraction}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles} \mathbf{of} \mathbf{One} \mathbf{Component}}{\mathbf{Number} \mathbf{of} \mathbf{Moles} \mathbf{of} \mathbf{the} \mathbf{Solvent}\mathbf{+}\mathbf{Number} \mathbf{of} \mathbf{Moles} \mathbf{of} \mathbf{Solute}} \\ \mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{8974}}{\mathbf{2}\mathbf{.}\mathbf{774}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{8974}} \\ \mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{8974}}{\mathbf{3}\mathbf{.}\mathbf{67}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{245}. \end{gathered}$$