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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.

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    $\begingroup$ Please note: 1. The quantity ‘amount of substance’ shall not be called ‘number of moles’, just as the quantity ‘mass’ shall not be called ‘number of kilograms’. 2. Descriptive terms or names of quantities shall not be arranged in the form of an equation. 3. Multiletter abbreviated terms (such as ‘MM’) shall not be used in the place of symbols. $\endgroup$ – Loong Jan 20 at 17:40
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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.

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  • $\begingroup$ Thank you. To be honest, I avoid memorizing too much formula. What confuses me though (until now) is the line "2.00m solution of NH3 in water". How did you know that there are 2 "moles" of NH3? Since that "2" from the question is the molal solution or molality of ammonia = 2 and its unit is mol/kg which is not the same with the number of moles (n), which is just "mol". Sorry for such question, I'm new to this. $\endgroup$ – Jayce Jan 21 at 5:07
  • $\begingroup$ @Jayce - The problem is open ended, so one can assume as much solution as desired. Frankly I tried solving the problem as 2 molar ( ie 1 liter of solution) which gave the "wrong" answer. Then I tried 2 molal (ie 1 kg of solvent) and got the "right" answer. An old convention is to use M for molar and m for molal. But without knowing which convention the particular book is using, that is somewhat of a guess. I think the newer convention is to be more explicit and use mol/L and mol/kg . $\endgroup$ – MaxW Jan 21 at 7:41
  • $\begingroup$ @Jayce - I edited the solution and moved things around a bit. Does that make the line of thought clearer? $\endgroup$ – MaxW Jan 21 at 8:33
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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.
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  • $\begingroup$ Thank you. I would like to ask some questions though. 1. Xi stands for mole fraction of i-th component, so if for example I was asked to find the mole fraction of the solvent, instead of the solute, will the formula be the same? 2. The reason for expressing the molality in mol kg^-1 is so that it will have the same unit as the molar mass of the solvent? 3. This is too much to ask but can you answer the problem by using the formulas I've written above (if it's possible). Or at least how to transform / derive it into your kinda shortcut formula. Again, thank you~ $\endgroup$ – Jayce Jan 20 at 13:21
  • $\begingroup$ 1. Yes, with respect to molar mass of solute, or just use $x_\mathrm{solv}=1-x_i$ for a single dissolved component; 2. No, 1 molal solution is a solution of 1 mol of the given compound in 1 kg of solvent by definition (not related to molar mass at all); 3. Since you used non-standard notations (or none at all) I'd rather prefer not to do that as it's going to bring a lot of confusion on both sides; I'll try to post an updated answer with the derivation later this day. $\endgroup$ – andselisk Jan 20 at 13:28
  • $\begingroup$ @Jayce The answer is updated with the derivation of the formula linking molality with mole fraction $\endgroup$ – andselisk Jan 20 at 13:53
  • $\begingroup$ Thank you again. It is clear now how the formula was derived. One of the reason I got too confused in answering the problem was due to the line in question: "2.00m solution of NH3". I assumed that the 2 molal is the molality of ammonia and not the solvent / water. Another reason was I kept on figuring out how can I insert the molar mass of NH3 into the formula and also how can I find the mass of water and ammonia given the limited givens. Again, thank you. I learned a new formula, thanks to you~ $\endgroup$ – Jayce Jan 20 at 14:49
  • $\begingroup$ @Jayce No prob, and good luck with chemistry:) $\endgroup$ – andselisk Jan 20 at 14:49

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