Question 13.2: Converting Molarity to Mole Fraction Laboratory ammonia is 1...
Converting Molarity to Mole Fraction
Laboratory ammonia is 14.8 M NH_3(aq) with a density of 0.8980 g/mL. What is x_{NH_3} in this solution?
Analyze
In this problem we note that no volume of solution is stated, suggesting that our calculation can be based on any fixed volume of our choice. A convenient volume to work with is one liter. We need to determine the number of moles of NH_3 and of H_2O in one liter of the solution.
Find the number of moles of NH_3 by using the definition of molarity.
\text{moles of } NH_3 = 1.00 L \times \frac{14.8 mol NH_3}{1 L} = 14.8 mol NH_3
For moles of H_2O, first find mass of the solution by using solution density.
\text { mass of soln } = 1000.0 mL \text { soln } \times \frac{0.8980 g \text { soln }}{1.0 mL \text { solution }} = 898.0 g \text { soln }
Then use moles of NH_3 and molar mass to find the mass of NH_3.
\text { mass of } NH_3 = 14.8 mol NH_3 \times \frac{17.03 g NH _3}{1 mol NH_3 } = 252 g NH_3
Find the mass of H_2O by subtracting the mass of NH_3 from the solution mass.
\text { mass of } H_2 O = 898.0 g \text { soln } – 252 g NH_3 = 646 g H_2 O
Find moles of H_2O by multiplying by the inverse of the molar mass for H_2O.
\text { moles of } H_2 O = 646 g H_2 O \times \frac{1 mol H_2 O }{18.02 g H _2 O } = 35.8 mol H_2 O
Find the mole fraction of ammonia x_{NH_3} by dividing moles NH_3 by the total number of moles of NH_3 and H_2O in the solution.
x_{NH_3} = \frac{14.8 mol NH_3}{14.8 mol NH_3 + 35.8 mol H_2O} = 0.292
Assess
By using the solution concentration definitions, we were able to convert from one concentration unit to another. This skill is used frequently by chemists.