
Find out the mole fraction of solute in an aqueous solution of 30% NaOH.
Answer
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Hint: As we know that mole fraction is the way of expressing the concentration of a solution and can be described as the number of molecules or moles of a particular component in a mixture divided by the total number of moles in the given mixture. So here we have to calculate the mole fraction of solute in an aqueous solution of 30% NaOH.
Formula used:
We will use the following formula in this solution:-
${{x}_{i}}=\dfrac{{{n}_{i}}}{{{n}_{\text{total}}}}$
where,
${{x}_{i}}$ = mole fraction
${{n}_{i}}$ = number of moles of one component of solution
${{n}_{\text{total}}}$ = Total number of moles of solution
Complete answer:
Let us first begin with the calculation of mass of NaOH present in the solution followed by calculation of moles and mole fraction as follows:-
-Calculation of mass of NaOH present in an aqueous solution of 30% NaOH:-
Let us assume that the mass of the given aqueous solution is 100 grams.
So the mass of NaOH present in it = 30 grams
This means the mass of water in this solution = 70 grams.
-Calculation of number of moles of NaOH and water in the solution:-
Moles can be calculated as: $\text{n=}\dfrac{\text{Given mass}}{\text{Molar mass}}$
Molar mass of NaOH = (23 +16 + 1) g/mol = 40g/mol
Number of moles of NaOH = ${{\text{n}}_{NaOH}}\text{=}\dfrac{30g}{40g/mol}=0.75moles$
Molar mass of ${{H}_{2}}O$ = (2 + 16) g/mol = 18g/mol
Number of moles of ${{H}_{2}}O$= ${{\text{n}}_{{{H}_{2}}O}}\text{=}\dfrac{70g}{18g/mol}=3.89moles$
-Calculation of mole fraction of NaOH present in the solution:-
Let us substitute all the values in ${{x}_{i}}=\dfrac{{{n}_{i}}}{{{n}_{\text{total}}}}$:-
$\begin{align}
& \Rightarrow {{x}_{NaOH}}=\dfrac{{{n}_{NaOH}}}{{{n}_{\text{total}}}} \\
& \Rightarrow {{x}_{NaOH}}=\dfrac{0.75moles}{0.75moles+3.89moles} \\
& \Rightarrow {{x}_{NaOH}}=0.16 \\
\end{align}$
Hence, the mole fraction of solute in an aqueous solution of 30% NaOH is 0.16.
Note:
-Remember to use the units of each given value along with it so as to easily cancel the terms with others in order to obtain an accurate answer.
-Also moles fraction is a unitless quantity.
Formula used:
We will use the following formula in this solution:-
${{x}_{i}}=\dfrac{{{n}_{i}}}{{{n}_{\text{total}}}}$
where,
${{x}_{i}}$ = mole fraction
${{n}_{i}}$ = number of moles of one component of solution
${{n}_{\text{total}}}$ = Total number of moles of solution
Complete answer:
Let us first begin with the calculation of mass of NaOH present in the solution followed by calculation of moles and mole fraction as follows:-
-Calculation of mass of NaOH present in an aqueous solution of 30% NaOH:-
Let us assume that the mass of the given aqueous solution is 100 grams.
So the mass of NaOH present in it = 30 grams
This means the mass of water in this solution = 70 grams.
-Calculation of number of moles of NaOH and water in the solution:-
Moles can be calculated as: $\text{n=}\dfrac{\text{Given mass}}{\text{Molar mass}}$
Molar mass of NaOH = (23 +16 + 1) g/mol = 40g/mol
Number of moles of NaOH = ${{\text{n}}_{NaOH}}\text{=}\dfrac{30g}{40g/mol}=0.75moles$
Molar mass of ${{H}_{2}}O$ = (2 + 16) g/mol = 18g/mol
Number of moles of ${{H}_{2}}O$= ${{\text{n}}_{{{H}_{2}}O}}\text{=}\dfrac{70g}{18g/mol}=3.89moles$
-Calculation of mole fraction of NaOH present in the solution:-
Let us substitute all the values in ${{x}_{i}}=\dfrac{{{n}_{i}}}{{{n}_{\text{total}}}}$:-
$\begin{align}
& \Rightarrow {{x}_{NaOH}}=\dfrac{{{n}_{NaOH}}}{{{n}_{\text{total}}}} \\
& \Rightarrow {{x}_{NaOH}}=\dfrac{0.75moles}{0.75moles+3.89moles} \\
& \Rightarrow {{x}_{NaOH}}=0.16 \\
\end{align}$
Hence, the mole fraction of solute in an aqueous solution of 30% NaOH is 0.16.
Note:
-Remember to use the units of each given value along with it so as to easily cancel the terms with others in order to obtain an accurate answer.
-Also moles fraction is a unitless quantity.
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