Finding mole fraction from molality

Question

Calculate the mole fraction of ammonia in a $\pu{2.00 m}$ solution of $\ce{NH3}$ in water.

What I know is that the formula for mole fraction is

$$X = \frac{\text{no.-of-moles-of-solute}}{\text{(no.-of-moles-of-solute)} + \text{(no.-of-moles-of-solvent)}}$$

The solute is ammonia which is $\ce{NH3}$ with a molar mass (MM) of $\pu{17 g mol-1}$, whereas the solvent is water or $\ce{H2O}$ which has a molar mass of $\pu{18 g mol-1}$.

The $\pu{2.00 m}$ from the problem signifies molality (because of the small $\pu{m}$), and molality is

$$\frac{\text{no.-of-moles-of-solute}}{\text{mass-of-solvent-in-kg}}$$

With no. of moles

$$n = \frac{m}{\text{MM}}$$

Despite knowing the formulas, I can't seem to solve for the answer. The answer should be $0.0347$, but I can't seem to get the right solution.

Any help would be appreciated.

Answer 1

Despite the unconventional notations, your formula is generally correct; however, you should've express mole fraction via molality explicitly and only then plug in the numbers. By definition, mole fraction of $i$-th component $x_i$ is

$$x_i = \frac{n_i}{n_\mathrm{tot}}$$

where $n_i$ – amount of $i$-th component; $n_i$ – total amount of all mixture constituents. For a simple solution of a single component the following hold true:

$$x_i = \frac{n_i}{n_i + n_\mathrm{solv}}$$

where $n_\mathrm{solv}$ – amount of the solvent that can also be found via its molecular mass $M_\mathrm{solv}$ and mass $m_\mathrm{solv}$, which, in turn, appears in the expression for molarity $b_i$:

$$b_i = \frac{n_i}{m_\mathrm{solv}} \quad\implies\quad m_\mathrm{solv} = \frac{n_i}{b_i}$$

$$n_\mathrm{solv} = \frac{m_\mathrm{solv}}{M_\mathrm{solv}} = \frac{n_i}{b_iM_\mathrm{solv}}$$

Finally, mole fraction can be expressed via molality as follows:

$$ \require{cancel} x_i = \frac{n_i}{n_i + n_\mathrm{solv}} = \frac{n_i}{n_i + \frac{n_i}{b_iM_\mathrm{solv}}} = \frac{\cancel{n_i}}{\cancel{n_i}\left(1 + (b_iM_\mathrm{solv})^{-1}\right)} = \frac{1}{1 + (b_iM_\mathrm{solv})^{-1}} $$

Time to plug in the numbers:

$$ \begin{align} x_i &= \frac{1}{1 + (b_iM_\mathrm{solv})^{-1}}\\ &= \frac{1}{1 + (\pu{2.00e-3 mol g-1}\cdot\pu{18.02 g mol-1})^{-1}}\\ &\approx 0.0347 \end{align} $$

Few key points:

1. Note that you have to convert molality expressed in $\pu{mol \color{red}{kg}-1}$ before plugging in the value: $$\pu{1 m} = \pu{1 mol kg-1} = \pu{1e-3 mol g-1}$$
2. In general, never omit units in your calculations and use standardized notations.
3. Mind significant figures. Since molality is given with two decimal points, you also should've taken molecular mass with higher precision.

Answer 2

You don't have to memorize some weird formula like andselisk has proposed.

You have sufficient information to solve the problem:

Calculate the mole fraction of ammonia in a $\pu{2.00 molal}$ solution of $\ce{NH3}$ in water.

We can assume any quantity of solution, so let's assume 1.00 kg of solvent. So the mass of solvent (water) is $\pu{1 kilogram} = \pu{1000 g}$ in a molal solution by definition.

moles of water = $\dfrac{1000}{18.015} = 55.402$

For 1.00 kg of solvent there are 2 moles of $\ce{NH3}$ which has a mass of $\pu{2 moles}\times \pu{17.031 g/mole} = \pu{34.062 g}$

From the Op's formula:

$X = \frac{\text{no.-of-moles-of-solute}}{\text{(no.-of-moles-of-solute)} + \text{(no.-of-moles-of-solvent)}} = \dfrac{2}{2 + 55.402} \approx 0.0348$

Now I'll confess that the significant figures in this problem bother me. To have three significant figures the molality should have been given as 2.00 molal, not 2 molal.