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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.

First there are 2 molesWe can assume any quantity of $\ce{NH3}$ which has a masssolution, so let's assume 1.00 kg of $\pu{2 moles}\times \pu{17.031 g/mole} = \pu{34.062 g}$

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.

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

You have sufficient information to the problem:

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

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

So the mass of solvent (water) is $\pu{1 kilogram} = \pu{1000 g}$ in a molal solution by definition.

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

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.

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

You have sufficient information to the problem:

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

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

So the mass of solvent (water) is $\pu{1 kilogram} = \pu{1000 g}$ in a molal solution by definition.

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

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.