Question

The molarity of $N{H_3}$ in aqueous solution is 11.8 $mol \cdot d{m^{ - 3}}$ . Calculate the mole fraction of $N{H_3}$ in solution. The density of the solution is 0.916 $gm{\text{ }}c{m^{ - 3}}$.

Answer 1

The formula to find the mole fraction is
${\text{Mole fraction of substance = }}\dfrac{{{\text{Moles of substance }}}}{{{\text{Total number of moles present in solution}}}}$

Complete Step-by-Step Solution:
We will first find the numbers of moles of $N{H_3}$ and water present in the solution. Then, we can find mole fraction using their values.
- The molarity of $N{H_3}$ solution is given as 11.8 $mol{\text{ d}}{{\text{m}}^{ - 3}}$.
We know that 1 $mol \cdot d{m^{ - 3}}$ = 0.001 $mol \cdot c{m^{ - 3}}$
So, 11.8 $mol \cdot d{m^{ - 3}}$ = $11.8 \times 0.001 \times 1$ = 0.0118 $mol \cdot c{m^{ - 3}}$
- We know that molarity is the number of moles of solute present in a given volume. So, here, 0.0118 moles of $N{H_3}$ is present in 1 $c{m^{ - 3}}$ volume.
Molecular weight of $N{H_3}$ = Atomic weight of N + 3(Atomic weight of H) = 14+3(1) = 17$gmmo{l^{ - 1}}$.
We know that
${\text{Number of moles = }}\dfrac{{{\text{Weight}}}}{{{\text{Molecular weight}}}}{\text{ }}.....{\text{(1)}}$
So, for $N{H_3}$, we can write that
$0.0118 = \dfrac{{Weight}}{{17}}$
So, Weight = $0.0118 \times 17$ = 0.2006 gm
Now, we also know that
${\text{Density = }}\dfrac{{{\text{Weight}}}}{{{\text{Volume}}}}$
Density of the solution is given 0.916 $gm \cdot c{m^{ - 3}}$.
So, we can say that in 1 $c{m^{ - 3}}$ volume corresponds to 0.916 gm of solution.
Now, we can find the weight of water in 1 $c{m^{ - 3}}$ by following the formula.
Weight of water = Weight of solution – Weight of ammonia
Weight of water = 0.916 – 0.2006 = 0.7154 gm
Now, we can find the moles of water present by using equation (1). We will get
${\text{Number of moles = }}\dfrac{{0.7154}}{{18}} = 0.03974$
Now, we will find the mole fraction of $N{H_3}$ in the solution by following the formula.
${\text{Mole fraction of substance = }}\dfrac{{{\text{Moles of substance }}}}{{{\text{Total number of moles present in solution}}}}$

So, for ammonia, we can write that
${\text{Mole fraction of N}}{{\text{H}}_3}{\text{ = }}\dfrac{{0.0118}}{{0.0118 + 0.03974}} = \dfrac{{0.0118}}{{0.0515}} = 0.2291$

Note: Make sure that you remember the relation between the different units of volume. Note that the sum of mole fractions of all the components present in the solutions is always 1. Mole fraction of a component in a solution is always less than one.