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Chapter 13: Problem 52

Commercial concentrated aqueous ammonia is \(28 \% \mathrm{NH}_{3}\) by mass and has a density of \(0.90 \mathrm{~g} / \mathrm{mL}\). What is the molarity of this solution?

Short Answer

Expert verified
The molarity of the concentrated aqueous ammonia solution can be calculated as follows: 1. Mass of solution = \(0.90 \frac{\mathrm{g}}{\mathrm{mL}}\) 脳 1000 \(\mathrm{mL}\) = 900 g 2. Mass of \(\mathrm{NH}_{3}\) = 900 g 脳 28% = 252 g 3. Moles of \(\mathrm{NH}_{3}\) = \(\frac{252 \mathrm{~g}}{17 \mathrm{~g/mol}}\) = 14.82 moles 4. Molarity = \(\frac{14.82 \text{ moles}}{1 \text{ L}}\) = 14.82 M Therefore, the molarity of the concentrated aqueous ammonia solution is 14.82 M.

Step by step solution

01

Calculate the mass of \(\mathrm{NH}_{3}\) in 1 L of solution.

In order to accomplish this, we first need to find the mass of 1 L (1000 mL) of the solution. We can do this by multiplying the density of the solution by its volume: Mass of solution = Density 脳 Volume Mass of solution = \(0.90 \frac{\mathrm{g}}{\mathrm{mL}}\) 脳 1000 \(\mathrm{mL}\) Next, we use the percentage composition of the solution to find the mass of \(\mathrm{NH}_{3}\) in it. To do this, we multiply the mass of the solution by the mass percentage of \(\mathrm{NH}_{3}\): Mass of \(\mathrm{NH}_{3}\) = Mass of solution 脳 mass percentage of \(\mathrm{NH}_{3}\) Mass of \(\mathrm{NH}_{3}\) = Mass of solution 脳 28%
02

Convert the mass of \(\mathrm{NH}_{3}\) to moles.

To convert the mass of \(\mathrm{NH}_{3}\) to moles, we divide it by its molar mass: Moles of \(\mathrm{NH}_{3}\) = \(\frac{\text{Mass of }\mathrm{NH}_{3}}{\text{Molar mass of }\mathrm{NH}_{3}}\) Molar mass of \(\mathrm{NH}_{3}\) = 14 (N) + 3 脳 1 (H) = 17 \(\mathrm{g/mol}\)
03

Calculate the molarity of the solution.

To find the molarity, we divide the moles of \(\mathrm{NH}_{3}\) by the volume of the solution in liters (since we are calculating the mass of the solution for 1 L): Molarity = \(\frac{\text{Moles of }\mathrm{NH}_{3}}{\text{Volume of solution}}\) Now, we can combine and calculate the steps above to find the molarity of the concentrated aqueous ammonia solution.

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Most popular questions from this chapter

The Henry's law constant for helium gas in water at \(30^{\circ} \mathrm{C}\) is \(3.7 \times 10^{-4} \mathrm{M} / \mathrm{atm}\) and the constant for \(\mathrm{N}_{2}\) at \(30{ }^{\circ} \mathrm{C}\) is \(6.0 \times 10^{-4} \mathrm{M} / \mathrm{atm} .\) If the two gases are each present at \(1.5\) atm pressure, calculate the solubility of each gas.

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Seawater contains \(3.4 \mathrm{~g}\) of salts for every liter of solution. Assuming that the solute consists entirely of \(\mathrm{NaCl}\) (over \(90 \%\) is), calculate the osmotic pressure of seawater at \(20^{\circ} \mathrm{C}\).

(a) What is the molality of a solution formed by dissolving \(1.25\) mol of \(\mathrm{KCl}\) in \(16.0\) mol of water? (b) How many grams of sulfur \(\left(\mathrm{S}_{8}\right)\) must be dissolved in \(100.0 \mathrm{~g}\) naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) to make a \(0.12 \mathrm{~m}\) solution?

At \(20^{\circ} \mathrm{C}\) the vapor pressure of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is 75 torr, and that of toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right)\) is 22 torr. Assume that benzene and toluene form an ideal solution. (a) What is the composition in mole fractions of a solution that has a vapor pressure of 35 torr at \(20^{\circ} \mathrm{C} ?\) (b) What is the mole fraction of benzene in the vapor above the solution described in part (a)?
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