The molality (m) of a solution is referred to as the measurement of the moles of solute divided by the solvent (either in liters or kilogram).
Mathematically;
[tex]\mathbf{Molality = \dfrac{\text{moles of solute}}{volume\ of \ solvent}}[/tex]
However, the mole fraction is the ratio of a particular substance molecule inside a mixture related to the overall number of moles of all the substances.
For a given mixture of C6H6 and CCl4
Their mole fraction can be expressed as:
[tex]X_{C6H6} = \dfrac{mole \ of \ C_6H_6}{mole \ of \ C_6H_6 + mole \ of \ CCl_4}[/tex]
[tex]X_{CCl_4} = \dfrac{mole \ of \ CCl_4}{mole \ of \ C_6H_6 + mole \ of \ CCl_4}[/tex]
where;
- mole fraction of C6H6 = 0.7
- mole fraction of CCl4 = ???
We know that:
[tex]\mathbf{X_{C6H6} + X_{CCl_4} = 1}[/tex]
[tex]\mathbf{ X_{CCl_4} = 1-X_{C6H6} }[/tex]
[tex]\mathbf{ X_{CCl_4} = 1-0.7}[/tex]
[tex]\mathbf{ X_{CCl_4} =0.3}[/tex]
∴
[tex]0.3 = \dfrac{mole \ of \ CCl_4}{mole \ of \ C_6H_6 + mole \ of \ CCl_4}[/tex]
where;
n represents the number of moles;
[tex]\mathbf{0.3 = \dfrac{n_{ CCl_4}}{n_{C_6H_6 }+ n_{ CCl_4}}}[/tex]
[tex]\mathbf{0.3n_{C_6H_6 }+ 0.3 n_{ CCl_4} = n_{ CCl_4}}[/tex]
[tex]\mathbf{0.3n_{C_6H_6 }= n_{ CCl_4}-0.3 n_{ CCl_4} }[/tex]
[tex]\mathbf{0.3n_{C_6H_6 }= 0.7n_{ CCl_4}}[/tex]
[tex]\mathbf{n_{C_6H_6 }=\dfrac{ 0.7}{0.3}n_{ CCl_4}---(1)}[/tex]
Assuming, we have 1 Liter of the solution;
Then,
- the mass of the solution = 0.9 g/mol × 1000 ml
- mass of the solution = 900 g
Recall that:
mass of the solution = mass of the solute + mass of the solvent
where;
mass of the solute = no of moles of [tex]\mathbf{n{CCl_4}}[/tex] × molar mass of [tex]\mathbf{n{CCl_4}}[/tex]
mass of the solvent = no of moles of [tex]\mathbf{n{C_6H_6}}[/tex] × molar mass of [tex]\mathbf{n{C_6H_6}}[/tex]
∴
900 g = ([tex]\mathbf{n{C_6H_6}}[/tex] × 78) + ([tex]\mathbf{n{CCl_4}}[/tex] × 154)
From equation (1);
[tex]\mathbf{n_{C_6H_6 }=\dfrac{ 0.7}{0.3}n_{ CCl_4}---(1)}[/tex]
replacing the value of [tex]\mathbf{n{C_6H_6}}[/tex] into 900 g = ([tex]\mathbf{n{C_6H_6}}[/tex] × 78) + ([tex]\mathbf{n{CCl_4}}[/tex] × 154); we have:
[tex]900 g = (\mathbf{\dfrac{ 0.7}{0.3}n_{ CCl_4}}\times 78) + ( n_{ CCl_4}}\times 154)[/tex]
[tex]900 g = (\mathbf{\dfrac{ 7}{3}n_{ CCl_4}\times 78) + ( n_{ CCl_4}}\times 154)}[/tex]
[tex]900 g = (\mathbf {7}\times n_{ CCl_4}\times 26) + ( n_{ CCl_4}}\times 154)}[/tex]
[tex]\mathbf{900 g = (\mathbf {182 +154)}n_{ CCl_4}}[/tex]
[tex]\mathbf{900 g = (\mathbf {336)}n_{ CCl_4}}[/tex]
[tex]\mathbf{n_{ CCl_4}=\dfrac{900 g}{336}}[/tex]
[tex]\mathbf{n_{ CCl_4}=2.679}[/tex]
From equation (1)
[tex]\mathbf{n_{C_6H_6 }=\dfrac{ 0.7}{0.3}n_{ CCl_4}}[/tex]
[tex]\mathbf{n_{C_6H_6 }=\dfrac{ 0.7}{0.3}\times 2.679}[/tex]
[tex]\mathbf{n_{C_6H_6 }=6.25}[/tex]
Finally, the molality of the solution is:
[tex]\mathbf{Molality = \dfrac{\text{moles of solute}}{volume\ of \ solvent}}[/tex]
[tex]\mathbf{Molality = \dfrac{\text{2.679}}{900}\times 1000}[/tex]
Molality = 2.976
Molality ≅ 3.0 m
Therefore, we can conclude that the molality of a solution prepared by mixing C6H6 with CCl4 is 3.0 m.
Learn more about Molality here:
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