
What is the mole fraction of the solute in a \[1.00{\text{ }}m\] aqueous solution?
A) 0.03554
B) 0.0177
C) 0.177
D) 1.770
Answer
511k+ views
Hint: Mole fraction of a solute in a binary solution is the ratio of the number of moles of solute to the total number of moles of the solution. The total number of moles of the solution includes the number of moles of the solute and the number of moles of the solvent. Molality is the ratio of the number of moles of solute present in one kilogram of solvent.
Complete answer:
For example, consider a binary solution containing solute A and solvent A. Let \[{n_A}\] and \[{n_B}\] represent the number of moles of solute A and solvent B respectively. Also, let \[{\chi _A}\] and \[{\chi _B}\] represent the mole fractions of solute A and solvent B respectively. Then
\[
\Rightarrow {\chi _A} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}} \\
\Rightarrow {\chi _B} = \dfrac{{{n_B}}}{{{n_A} + {n_B}}} \\
\]
The molality of the solute is \[1.00{\text{ }}m\] .
One kilogram of solvent B contains one mole of solute A.
One kilogram is equal to one thousand grams.
\[1{\text{ kg = 1000 g}}\]
Thus, the mass of solvent is \[{\text{1000 g}}\]
In the aqueous solution, the solvent is water. Thus, the mass of water is \[{\text{1000 g}}\] .
The molecular weight of water is \[{\text{18 g/mol}}\] .
We will divide the mass of water with its molecular weight to obtain the number of moles of water.
\[
\Rightarrow {n_B} = \dfrac{{{\text{1000 g}}}}{{{\text{18 g/mol}}}} \\
\Rightarrow {n_B} = 55.56{\text{ mol}} \\
\]
Hence, the number of moles of water present is \[55.56{\text{ moles}}\] .
The number of moles of the solute A is one mole.
We will calculate the mole fraction of solute
\[
\Rightarrow {\chi _A} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}} \\
\Rightarrow {\chi _B} = \dfrac{1}{{1 + 55.56}} \\
\Rightarrow {\chi _B} = \dfrac{1}{{56.56}} \\
\Rightarrow {\chi _B} = 0.0177 \\
\]
Thus, the mole fraction of the solute in a \[1.00{\text{ }}m\] aqueous solution is 0.0177.
Hence, the correct answer is option B.
Note: A binary solution contains only two components. The component present in small amounts is the solute and the component present in large amounts is solvent. In a binary solution, the sum of the mole fractions of solute and solvent is equal to unity. \[{\chi _A} + {\chi _B} = 1\]
Complete answer:
For example, consider a binary solution containing solute A and solvent A. Let \[{n_A}\] and \[{n_B}\] represent the number of moles of solute A and solvent B respectively. Also, let \[{\chi _A}\] and \[{\chi _B}\] represent the mole fractions of solute A and solvent B respectively. Then
\[
\Rightarrow {\chi _A} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}} \\
\Rightarrow {\chi _B} = \dfrac{{{n_B}}}{{{n_A} + {n_B}}} \\
\]
The molality of the solute is \[1.00{\text{ }}m\] .
One kilogram of solvent B contains one mole of solute A.
One kilogram is equal to one thousand grams.
\[1{\text{ kg = 1000 g}}\]
Thus, the mass of solvent is \[{\text{1000 g}}\]
In the aqueous solution, the solvent is water. Thus, the mass of water is \[{\text{1000 g}}\] .
The molecular weight of water is \[{\text{18 g/mol}}\] .
We will divide the mass of water with its molecular weight to obtain the number of moles of water.
\[
\Rightarrow {n_B} = \dfrac{{{\text{1000 g}}}}{{{\text{18 g/mol}}}} \\
\Rightarrow {n_B} = 55.56{\text{ mol}} \\
\]
Hence, the number of moles of water present is \[55.56{\text{ moles}}\] .
The number of moles of the solute A is one mole.
We will calculate the mole fraction of solute
\[
\Rightarrow {\chi _A} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}} \\
\Rightarrow {\chi _B} = \dfrac{1}{{1 + 55.56}} \\
\Rightarrow {\chi _B} = \dfrac{1}{{56.56}} \\
\Rightarrow {\chi _B} = 0.0177 \\
\]
Thus, the mole fraction of the solute in a \[1.00{\text{ }}m\] aqueous solution is 0.0177.
Hence, the correct answer is option B.
Note: A binary solution contains only two components. The component present in small amounts is the solute and the component present in large amounts is solvent. In a binary solution, the sum of the mole fractions of solute and solvent is equal to unity. \[{\chi _A} + {\chi _B} = 1\]
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