A bottle of commercial concentrated aqueous ammonia is labeled "29.89\% \(\mathrm{NH}_{3}\) by mass; density \(=0.8960 \mathrm{~g} / \mathrm{mL}\)." (a) What is the molarity of the ammonia solution? (b) If \(250.0 \mathrm{~mL}\) of the commercial ammonia is diluted with water to make \(3.00 \mathrm{~L}\) of solution, what is the molarity of the diluted solution?

First, we need to find the mass of \(\mathrm{NH}_{3}\) in 1 L of solution. We are given that the mass percentage of \(\mathrm{NH}_{3}\) in the solution is \(29.89\%\). We also know the density of the solution is \(0.8960 \mathrm{~g} / \mathrm{mL}\), so its mass in 1 L is: \(1000\mathrm{~mL} \times 0.8960 \mathrm{~g} / \mathrm{mL} = 896\mathrm{~g}\). We can then find the mass of \(\mathrm{NH}_{3}\) in 1 L of solution: \(896\mathrm{~g} \times 29.89\% = 267.95\mathrm{~g}\). Now, we can convert the mass of \(\mathrm{NH}_{3}\) to moles using its molar mass, which is \(17.03\mathrm{~g/mol}\): \(\displaystyle\frac{267.95\mathrm{~g}}{17.03\mathrm{~g/mol}} = 15.74\mathrm{~mol}\).

Now that we have the moles of ammonia in 1 L of solution, we can calculate the molarity. The formula for molarity is: \(M = \displaystyle\frac{n}{V}\) Plugging our values, we have: \(M = \displaystyle\frac{15.74\mathrm{~mol}}{1\mathrm{~L}} = 15.74\, \mathrm{M}\). So, the molarity of the ammonia solution is \(15.74\, \mathrm{M}\).

In the second part of the problem, we are asked to find the molarity of the diluted solution after adding the concentrated ammonia to water. We know the volume of the commercial ammonia added is \(250.0\mathrm{~mL}\) and the final volume of the diluted solution is \(3.00\mathrm{~L}\). We can use the dilution formula to find the molarity of the diluted solution: \(M_{1}V_{1} = M_{2}V_{2}\) Our initial molarity \(M_{1}\) is \(15.74\, \mathrm{M}\), and the initial volume \(V_{1}\) is \(250.0\mathrm{~mL}\). The final volume \(V_{2}\) is \(3.00\mathrm{~L}\). We can plug in the values and solve for \(M_{2}\): \(15.74\mathrm{M} \times 250.0\mathrm{~mL} = M_{2} \times 3.00\mathrm{~L}\) We need to convert \(V_{1} = 250.0\mathrm{~mL}\) to liters to match the units with \(V_{2}\): \(V_{1} = 250.0\mathrm{~mL} \times \displaystyle\frac{1\mathrm{~L}}{1000\mathrm{~mL}} = 0.250\mathrm{~L}\) Now we can solve for \(M_{2}\): \(M_{evential} = \displaystyle\frac{15.74\mathrm{M} \times 0.250\mathrm{~L}}{3.00\mathrm{~L}} = 1.31\, \mathrm{M}\). Therefore, the molarity of the diluted solution is \(1.31\, \mathrm{M}\).