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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 approximately 14.8 M.

Step by step solution

01

Formula for molarity

Molarity (M) is defined as the number of moles of solute (in this case, NH3) present in 1 liter of the solution. The formula for molarity is: \[M = \frac{moles \ of \ solute}{volume \ of \ solution \ (L)}\] #Step 2: Calculate the mass of NH3 in 1 mL of the solution#
02

Mass of NH3 in 1 mL

We are given that the concentrated ammonia solution is 28% NH3 by mass. Therefore, the mass of NH3 in 1 mL solution can be calculated as: Mass of NH3 in 1 mL = (percentage composition) × (mass of 1 mL solution) We are given that the density of the solution is 0.90 g/mL, so the mass of 1 mL solution = 0.90 g Mass of NH3 = 0.28 × 0.90 g = 0.252 g #Step 3: Calculate the moles of NH3 in 1 mL of the solution#
03

Moles of NH3 in 1 mL

We are given the mass of NH3 in 1 mL of the solution as 0.252 g. To convert this to moles, we will need to use the molecular weight of NH3. NH3 has one nitrogen atom (14 g/mol), and three hydrogen atoms (3 × 1 g/mol), so the molecular weight of NH3 is (14 + 3) g/mol = 17 g/mol. The moles of NH3 in 1 mL of the solution is: Moles of NH3 = mass of NH3 ÷ molar mass of NH3 Moles of NH3 = 0.252 g ÷ 17 g/mol ≈ 0.0148 mol #Step 4: Calculate the molarity of NH3 in the solution#
04

Molarity of NH3

Now that we know the moles of NH3 in 1 mL of the solution, we can convert this to the molarity of NH3 in the solution. Recall the formula for molarity: \[M = \frac{moles \ of \ solute}{volume \ of \ solution \ (L)}\] To find the molarity of the solution, we can divide the moles of NH3 in 1 mL of the solution by the volume of 1 mL in liters (0.001 L): Molarity = 0.0148 mol ÷ 0.001 L ≈ 14.8 M #Solution# The molarity of the commercial concentrated aqueous ammonia solution is approximately 14.8 M.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ammonia Solution
Ammonia solution is a mixture of ammonia ( H_3 ) dissolved in water. It's often used in cleaning products and industrial applications. The concentration of ammonia in solution is crucial for its effectiveness. Commercial ammonia solutions are often concentrated, containing a high percentage of ammonia by mass. Understanding how to calculate the concentration or molarity is important for both practical applications and academic studies. In this context, let's explore how the components like percentage composition and density help determine the molarity of the solution.
Percentage Composition
Percentage composition refers to the mass percentage of a particular component in a solution. In the case of our ammonia solution, it's given as 28%, meaning 28 grams of ammonia are present in every 100 grams of solution. This measurement helps us determine how much of the solute (ammonia) is in the solution compared to the total mass. Knowing the percentage composition allows us to calculate the exact mass of ammonia in a specific volume, leading to more precise molarity calculations.
Density
Density is an important property that links mass and volume. It is defined as mass per unit volume and expressed in \( ext{g/mL} \). For our ammonia solution, the density is given as 0.90 \( ext{g/mL} \). This means every milliliter of solution has a mass of 0.90 grams. By knowing the density, we can easily convert between mass and volume, which is crucial when calculating the moles of a solute in a given volume of solution. Understanding density allows us to determine how much mass of ammonia is contained in a certain volume, aiding in the calculation of moles.
Molecular Weight
Molecular weight, also known as molar mass, is the mass of one mole of a substance. For ammonia (H_3), it consists of one nitrogen atom and three hydrogen atoms, leading to a molar mass of 17 \( ext{g/mol} \). Understanding molecular weight is essential when converting mass to moles. It's calculated as the sum of the atomic masses of all atoms in a molecule. This concept is fundamental in chemistry, particularly when dealing with solution concentrations and reactions. It connects the concept of mass to the amount of substance present.
Moles Calculation
Moles represent a quantity that chemists use to count particles like atoms and molecules. The calculation of moles from mass involves dividing the mass of a substance by its molar mass. Using the ammonia solution example, once we know the mass of H_3 in a given volume of solution, we divide that mass by the molecular weight, 17 \( ext{g/mol} \), to find the number of moles. This calculation is necessary for finding molarity, which is the moles of solute per liter of solution. By mastering moles calculation, students can better understand solution concentrations and chemical reactions.

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Most popular questions from this chapter

Oil and water are immiscible. Which is the most likely reason? (a) Oil molecules are denser than water. (b) Oil molecules are composed mostly of carbon and hydrogen. (c) Oil molecules have higher molar masses than water. (d) Oil molecules have higher vapor pressures than water. (e) Oil molecules have higher boiling points than water.

When ammonium chloride dissolves in water, the solution becomes colder. (a) Is the solution process exothermic or endothermic? (b) Why does the solution form?

Which of the following in each pair is likely to be more soluble in water: (a) cyclohexane (C. \(\mathrm{H}_{12}\) ) or glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) (b) propionic acid \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right)\) or sodium propionate \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COONa}\right),(\mathbf{c}) \mathrm{HCl}\) or ethyl chloride \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\right) ?\) Explain in each case.

At \(35^{\circ} \mathrm{C}\) the vapor pressure of acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO},\) is 360 torr, and that of chloroform, \(\mathrm{CHCl}_{3},\) is 300 torr. Acetone and chloroform can form very weak hydrogen bonds between one another; the chlorines on the carbon give the carbon a sufficient partial positive charge to enable this behavior: A solution composed of an equal number of moles of acetone and chloroform has a vapor pressure of 250 torr at \(35^{\circ} \mathrm{C}\) (a) What would be the vapor pressure of the solution if it exhibited ideal behavior? (b) Based on the behavior of the solution, predict whether the mixing of acetone and chloroform is an exothermic \(\left(\Delta H_{\text { soln }}<0\right)\) or endothermic \(\left(\Delta H_{\text { soln }}>0\right)\) process.

Carbon disulfide \(\left(\mathrm{CS}_{2}\right)\) boils at \(46.30^{\circ} \mathrm{C}\) and has a density of 1.261 \(\mathrm{g} / \mathrm{mL}\) . (a) When 0.250 \(\mathrm{mol}\) of a nondissociating solute is dissolved in 400.0 \(\mathrm{mL}\) of \(\mathrm{CS}_{2},\) the solution boils at \(47.46^{\circ} \mathrm{C} .\) What is the molal boiling-point-elevation constant for \(\mathrm{CS}_{2}\) ? (b) When 5.39 \(\mathrm{g}\) of a nondissociating unknown is dissolved in 50.0 \(\mathrm{mL}\) of \(\mathrm{CS}_{2},\) the solution boils at \(47.08^{\circ} \mathrm{C} .\) What is the molar mass of the unknown?

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