Question: The mole fraction of \[HN{O_3}\] in an aqueous binary solution is \[0.15\]. The ratio of moles of \[...

Question

The mole fraction of $HN{O_3}$ in an aqueous binary solution is $0.15$ . The ratio of moles of $HN{O_3}$ to ${H_2}O$ in the solution is nearly:
A) 17:3
B) 3:17
C) 1:1
D) 15:100

Answer 1

Solution

Mole fraction represents the number of molecules of a particular component in a mixture divided by the total number of moles in the given mixture. It’s a way of expressing the concentration of a solution.
The molar fraction can be represented by X. If the solution consists of components A and B, then the mole fraction is,
Mole fraction of solute $= \dfrac{{Mole{\text{ }}of{\text{ }}solute}}{{Mole{\text{ }}of{\text{ }}solute + Moles{\text{ }}of{\text{ }}solvant}}$
Therefore, the sum of mole fraction of all the components is always equal to one.
Go by the definition of mole fraction and divide by moles of water to get the value of required ratio, which is the number of moles of $HN{O_3}$ to the number of moles of ${H_2}O$ .

Complete step by step answer:
Mole-fraction is defined as the ratio of the number of moles of a component of a solution to the total number of moles of all components. For a two component (binary) solution having components A and B
Mole fraction of component A $= \dfrac{{{n_A}}}{{{n_A} + {n_B}}}$
where, ${n_A}$ is the number of moles of A and ${n_B}$ is the number of moles of B
In the present situation, there is a binary solution, it has only two components, $HN{O_3}$ and ${H_2}O$ .
Therefore, by the definition of mole-fraction:
$\dfrac{{{n_{HN{O_3}}}}}{{{n_{HN{O_3}}} + {n_{{H_2}O}}}} = 0.15$
Now, we require $\dfrac{{{n_{HN{O_3}}}}}{{{n_{{H_2}O}}}}$
So, divide numerator and denominator of the left-hand-side by ${n_{{H_2}O}}$
$= > \dfrac{{\dfrac{{{n_{HN{O_3}}}}}{{{n_{{H_2}O}}}}}}{{\dfrac{{{n_{HN{O_3}}} + {n_{{H_2}O}}}}{{{n_{{H_2}O}}}}}} = \dfrac{{15}}{{100}}$
$= > \dfrac{{\dfrac{{{n_{HN{O_3}}}}}{{{n_{{H_2}O}}}}}}{{\dfrac{{{n_{HN{O_3}}}}}{{{n_{{H_2}O}}}} + 1}} = \dfrac{3}{{20}}$
Put, $\dfrac{{{n_{HN{O_3}}}}}{{{n_{HN{O_3}}} + {n_{{H_2}O}}}} = 0.15$ in the above equation. We get
$x = \dfrac{3}{{17}}$ which is the required ratio.

Therefore, the correct answer is (B).

Note: One should not go on to find individual moles to find the ratio of moles, which is impossible in this case. So, the required ratio of the number of moles should be found by dividing with the number of moles of water as shown above. The mole-fraction of $0.15$ should be converted to a rational number as the options are in the form of rational numbers and not decimals.