Question

Calculate the mole fractions, molarity and molality of \[HN{{O}_{3}}\] in a solution containing $12.2\%$ \[HN{{O}_{3}}\]. (Given density of \[HN{{O}_{3}}\] is 1.038 $g/c{{m}^{3}}$).

Answer 1

Hint: Mass of \[HN{{O}_{3}}\] can be calculated as density and volume are already given. Then calculate the number of moles by using the following formula:
\[No.\text{ of moles}=\dfrac{\operatorname{Given}Mass}{\operatorname{Molar}Mass}\]
After this, the mole fraction, molarity and molality can be easily calculated from their respective formulas.

Complete step-by-step answer:
Let’s first calculate the molar mass of \[HN{{O}_{3}}\]. We see single atoms of hydrogen and nitrogen with 3 atoms of oxygen. Adding all of that, the molar mass of nitric acid is 63 g.
It is given in the question that the solution contains $12.2\%$ of \[HN{{O}_{3}}\], which otherwise means, there is 12.2 ml of \[HN{{O}_{3}}\] in 100ml of the solution. 12.2 ml is equal to 12.2 $c{{m}^{3}}$ (as 1$c{{m}^{3}}$ is equal to 1ml). Using unitary method to solve this, we get,
\[\begin{align}
& 100\operatorname{ml}\to 12.2\operatorname{ml} \\
& 1\operatorname{ml}\to \dfrac{12.2}{100}=0.122ml \\
\end{align}\]
So, we found out that 1ml of the solution contains 0.122 $c{{m}^{3}}$ of nitric acid. Using the density given, we can calculate the mass of nitric acid present in 1ml of the solution, which is as shown below:
\[\begin{align}
& Density=\dfrac{Mass}{Volume} \\
& \Rightarrow Mass=Density\times Volume=1.038\times 0.122 \\
& \Rightarrow Mass=0.126636\operatorname{g} \\
\end{align}\]
The total mass of nitric acid present in 100ml of solution is:
\[0.126636\times 12.2=1.544\operatorname{g}\].
The number of moles of nitric acid in the solution is:
\[No.\text{ of moles}=\dfrac{\operatorname{Given}Mass}{\operatorname{Molar}Mass}=\dfrac{1.544}{63}=0.0245\operatorname{moles}\]
We are at a position to calculate the molarity of this solution. As you know, molarity is defined as the number of moles present in 1000 ml of the solution. Here 0.0245 moles of nitric acid is present in 100 ml of the solution. We again use unitary method as outlined below:
\[\begin{align}
 & 100\operatorname{ml}\to 0.0245\operatorname{moles} \\
 & 1000\operatorname{ml}\to 0.245\operatorname{moles} \\
\end{align}\]
Which implies the molarity of the solution is 0.245 M.
Let’s move on to the morality of the solution. Molality is defined as the number of moles of solute present in 1000 g of the solvent. Our very first step should be to convert everything into grams. Out of the 100 ml solution, we have 87.8 ml of water (which is the solvent here). The density of water is 1 $g/c{{m}^{3}}$ which means that 87.8 ml of water is equal to 87.8 g of solvent. Using unitary method, we calculate as:
\[\begin{align}
  & 87.8\operatorname{g}\text{ of solvent}\to 1.544\operatorname{g}\text{ of HN}{{\text{O}}_{3}} \\
 & 1000\operatorname{g}\text{ of solvent}\to \dfrac{1.544\times 1000}{87.8}=17.585\operatorname{g}\text{ of HN}{{\text{O}}_{3}} \\
\end{align}\]
Now we convert 17.585 g into moles by the following process:
\[No.\text{ of moles}=\dfrac{\operatorname{Given}Mass}{\operatorname{Molar}Mass}=\dfrac{17.585}{63}=0.279\operatorname{moles}\]
As the 0.279 moles of nitric acid is present in 1000 g of solvent, so the molality of the solution is 0.279 molal.
Mole fraction is defined as the ratio of number of moles of a particular constituent with that of the total number of moles present in the solution. We have already calculated that in 100ml of the given solution, nitric acid is 0.0245 moles. We have to calculate the moles of water present in the solution.
\[No.\text{ of moles of water}=\dfrac{\operatorname{Given}Mass}{\operatorname{Molar}Mass}=\dfrac{87.8}{18}=4.877\operatorname{moles}\]
So, the mole fraction of nitric acid is calculated as below:
\[Mole\text{ fraction of HN}{{\text{O}}_{3}}=\dfrac{0.0245}{0.0245+4.877}=0.00499\]

Therefore, the mole fraction of nitric acid in the solution is 0.00499.

Note: All the stoichiometric calculations done above, are very unit specific in nature. Students are advised to maintain caution during solving. Mole fraction is a dimensionless quantity because it is a ratio between the same quantities.