What is the molality of a solution prepared by mixing C6H6 with CCl4. If the solution density is 0.9 g/ml and the mole fraction of C6H6 in the solution is 0.7​

Answers

Answer 1

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 = 0.7mole 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 mlmass 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.

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It is given that, 2 g of suppositories is needed to prepare. The part of water in the whole of the given composition is 10. The 10% of 2 g is 0.2 g. Therefore, 0.2 g of water that is 0.2 ml of water is needed for the given recipe.

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