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Chapter 13: Problem 56

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 commercial concentrated aqueous ammonia solution is 14.82 M.

Step by step solution

01

Calculate the mass of ammonia in 1L of solution

: First, we need to find out the mass of ammonia (NH3) in 1L of the solution. Given that the density of the solution is 0.90 g/mL, we can determine the mass of 1L of the solution by multiplying the density by the volume: Mass of 1L solution = Density × Volume Mass of 1L solution = \(0.90 \frac{\mathrm{g}}{\mathrm{mL}} × 1000 \mathrm{mL}\) Mass of 1L solution = 900 g Now, we can calculate the mass of ammonia in 1L of the solution by using the given percentage by mass: Mass of ammonia = (Percentage by mass × Mass of 1L solution) / 100 Mass of ammonia = \(28\% × 900 \mathrm{g}\) / 100 Mass of ammonia = 252 g
02

Convert the mass of ammonia to moles

: Next, we need to convert the mass of ammonia to moles using its molar mass. The molar mass of NH3 is approximately 17 g/mol (14 g/mol for N and 3 g/mol for H). Moles of ammonia = Mass of ammonia / Molar mass of ammonia Moles of ammonia = \(252 \mathrm{g} / 17 \frac{\mathrm{g}}{\mathrm{mol}}\) Moles of ammonia = 14.82 moles
03

Calculate the molarity of the solution

: Finally, we can calculate the molarity of the ammonia solution using the moles of ammonia and the volume of the solution. Molarity = Moles of solute / Volume of solution (in Liters) Molarity = \(14.82 \mathrm{moles} / 1 \mathrm{L}\) Molarity = 14.82 M Therefore, the molarity of the commercial concentrated aqueous ammonia solution is 14.82 M.

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Most popular questions from this chapter

What is the molarity of each of the following solutions: (a) \(15.0 \mathrm{~g}\) of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in \(0.250 \mathrm{~mL}\) solution, (b) \(5.25 \mathrm{~g}\) of $\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\( in \)175 \mathrm{~mL}$ of solution, (c) \(35.0 \mathrm{~mL}\) of \(9.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) diluted to $0.500 \mathrm{~L} ?$

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Calculate the molarity of the following aqueous solutions: (a) \(0.640 \mathrm{~g}\) of \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) in \(500.0 \mathrm{~mL}\) of solution, (b) \(50.0 \mathrm{~g}\) of \(\mathrm{LiClO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}\) in $250 \mathrm{~mL}$ of solution, (c) \(125 \mathrm{~mL}\) of \(3.00 \mathrm{M}\) \(\mathrm{HNO}_{3}\) diluted to \(1.00 \mathrm{~L}\)

A solution contains \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) and an unknown number of moles of sodium chloride. The vapor pressure of the solution at \(29^{\circ} \mathrm{C}\) is \(3.85 \mathrm{kPa}\). The vapor pressure of pure water at this temperature is \(4.05 \mathrm{kPa}\). Calculate the number of grams of sodium chloride in the solution. (Hint: Remember that sodium chloride is a strong electrolyte.)
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