2 added 99 characters in body edited Jan 21 at 8:32 MaxW 16.7k22 gold badges2323 silver badges6464 bronze badges 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. 1 answered Jan 20 at 17:25 MaxW 16.7k22 gold badges2323 silver badges6464 bronze badges 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.