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Calculate the molarity of each of these solutions. a. A 5.623-g sample of \(\mathrm{NaHCO}_{3}\) is dissolved in enough water to make \(250.0 \mathrm{~mL}\) of solution. b. A \(184.6-\mathrm{mg}\) sample of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is dissolved in enough water to make \(500.0 \mathrm{~mL}\) of solution. c. A \(0.1025-\mathrm{g}\) sample of copper metal is dissolved in \(35 \mathrm{~mL}\) of concentrated \(\mathrm{HNO}_{3}\) to form \(\mathrm{Cu}^{2+}\) ions and then water is added to make a total volume of \(200.0 \mathrm{~mL}\). (Calculate the molarity of \(\mathrm{Cu}^{2+}\).)

Short Answer

Expert verified
The molarities of the solutions are as follows: a. \(0.26776\text{ M}\) for the NaHCO3 solution b. \(0.001255\text{ M}\) for the K2Cr2O7 solution c. \(0.00807\text{ M}\) for the Cu²⁺ ions in solution

Step by step solution

01

Find the moles of NaHCO3

First, we need to find the molar mass of NaHCO3. Molar mass of NaHCO3 = 22.99(g/mol for Na) + 1.01(g/mol for H) + 12.01(g/mol for C) + 3 * 16.00(g/mol for O) = 84.01 g/mol Now, we'll find the moles of NaHCO3 by dividing the mass by the molar mass: moles of NaHCO3 = 5.623 g / 84.01 g/mol = 0.06694 mol
02

Convert volume to liters

We need to convert the volume of the solution from mL to L: 250 mL = 250/1000 = 0.250 L
03

Calculate molarity

Now we can find the molarity using the formula: Molarity (M) = moles of solute / volume of solution in L Molarity (M) = 0.06694 mol / 0.250 L = 0.26776 M #b. Finding the molarity of K2Cr2O7 solution#
04

Find the moles of K2Cr2O7

First, find the molar mass of K2Cr2O7. Molar mass of K2Cr2O7 = 2 * 39.10(g/mol for K) + 2 * 51.996(g/mol for Cr) + 7 * 16.00(g/mol for O) = 294.18 g/mol Next, convert the mass from mg to g: 184.6 mg = 184.6 / 1000 = 0.1846 g Now, find the moles of K2Cr2O7 by dividing the mass by the molar mass: moles of K2Cr2O7 = 0.1846 g / 294.18 g/mol = 0.0006274 mol
05

Convert volume to liters

We need to convert the volume of the solution from mL to L: 500 mL = 500/1000 = 0.500 L
06

Calculate molarity

Now we can find the molarity using the formula: Molarity (M) = moles of solute / volume of solution in L Molarity (M) = 0.0006274 mol / 0.500 L = 0.001255 M #c. Finding the molarity of Cu²⁺ ions in solution#
07

Find the moles of Cu

First, find the molar mass of Cu. Molar mass of Cu = 63.546 g/mol Now, find the moles of Cu by dividing the mass by the molar mass: moles of Cu = 0.1025 g / 63.546 g/mol = 0.001614 mol
08

Convert volume to liters

We need to convert the total volume of the solution from mL to L: 200 mL = 200/1000 = 0.200 L
09

Calculate molarity

Now we can find the molarity using the formula: Molarity (M) = moles of solute / volume of solution in L Molarity (M) = 0.001614 mol / 0.200 L = 0.00807 M

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Key Concepts

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

Moles Conversion
Converting a given mass into moles is an essential step in calculating the molarity of a solution. This process involves using the substance's molar mass as a conversion factor. To begin, it's important to understand the relationship between moles, mass, and molar mass. The formula is:
  • \( ext{Moles} = \frac{\text{Mass in grams}}{\text{Molar Mass in g/mol}} \)
Once you have the mass of the solute, simply divide it by the molar mass of the substance. For instance, in the exercise, the moles of \(\mathrm{NaHCO}_3\) are calculated using its given mass of 5.623 grams and its molar mass. Dividing these provides the number of moles, a key value that leads us closer to finding the solution's molarity. Repeating this process for each solute ensures accurate molarity calculations.
Molar Mass Calculation
Before calculating the number of moles, you need to determine the molar mass of the component elements within the compound. This includes summing up the atomic masses of all the atoms present in the chemical formula. For example, consider \(\mathrm{NaHCO}_3\):
  • Sodium (Na): 22.99 g/mol
  • Hydrogen (H): 1.01 g/mol
  • Carbon (C): 12.01 g/mol
  • Oxygen (O): 16.00 g/mol; since there are three oxygen atoms, multiply by 3
The resulting calculation \(22.99 + 1.01 + 12.01 + 3\times16.00 = 84.01\) provides the molar mass.
Similarly, follow this procedure for substances like \(\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7\). Accurate molar mass calculation is crucial for the subsequent moles conversion.
Volume Conversion
Volume is often provided in milliliters (mL) but must be converted into liters (L) to calculate molarity. The conversion is simple: divide the volume in milliliters by 1000.
  • Volume in liters = Volume in milliliters / 1000
In our exercise example, converting 250 mL of \(\mathrm{NaHCO}_3\) solution into liters results in 0.250 L. This is a critical step because molarity, defined as moles per liter, requires the volume to be in standard liters.
Applying this conversion ensures uniformity and accuracy in solving for the molarity of any solution, such as \(\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7\) and \(\mathrm{Cu}^{2+}\) ions.
Molar Concentration
Molarity, or molar concentration, refers to the number of moles of solute per liter of solution. It is a fundamental concept in chemistry for expressing concentration. The formula for calculating molarity is:
  • \( ext{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \)
Utilizing this formula after converting both moles and volume becomes straightforward. Take, for example, the \(\mathrm{NaHCO}_3\) solution. With 0.06694 moles of \(\mathrm{NaHCO}_3\) in 0.250 liters of solution, the molarity computes to 0.26776 M.
Understanding how to apply these calculations offers valuable insights into the solution's concentration, which is essential for various practical chemical applications.

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Most popular questions from this chapter

A mixture contains only sodium chloride and potassium chloride. A \(0.1586-\mathrm{g}\) sample of the mixture was dissolved in water. It took \(22.90 \mathrm{~mL}\) of \(0.1000 \mathrm{M} \mathrm{AgNO}_{3}\) to completely precipitate all the chloride present. What is the composition (by mass percent) of the mixture?

The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$ \mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }} $$ In the case of very dilute aqueous solutions, a concentration of \(1.0\) ppm is equal to \(1.0 \mu \mathrm{g}\) of solute per \(1.0 \mathrm{~mL}\), which equals \(1.0 \mathrm{~g}\) solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. \(5.0 \mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. \(1.0 \mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. \(10.0 \mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. \(0.10 \mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)

A \(1.42-\mathrm{g}\) sample of a pure compound, with formula \(\mathrm{M}_{2} \mathrm{SO}_{4}\), was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected, dried, and found to weigh \(1.36 \mathrm{~g}\). Determine the atomic mass of \(\mathrm{M}\), and identify \(\mathrm{M}\).

Describe how you would prepare \(2.00 \mathrm{~L}\) of each of the following solutions. a. \(0.250 \mathrm{M} \mathrm{NaOH}\) from solid \(\mathrm{NaOH}\) b. \(0.250 M \mathrm{NaOH}\) from \(1.00 M \mathrm{NaOH}\) stock solution c. \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) from solid \(\mathrm{K}_{2} \mathrm{CrO}_{4}\) d. \(0.100 M \mathrm{~K}_{2} \mathrm{CrO}_{4}\) from \(1.75 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) stock solution

Consider reacting copper(II) sulfate with iron. Two possible reactions can occur, as represented by the following equations. \(\operatorname{copper}(\) II \()\) sulfate \((a q)+\) iron \((s) \longrightarrow\) \(\operatorname{copper}(s)+\) iron(II) sulfate \((a q)\) copper(II) sulfate \((a q)+\operatorname{iron}(s) \longrightarrow\) \(\operatorname{copper}(s)+\) iron(III) sulfate \((a q)\) You place \(87.7 \mathrm{~mL}\) of a \(0.500-M\) solution of copper(II) sulfate in a beaker. You then add \(2.00 \mathrm{~g}\) of iron filings to the copper(II) sulfate solution. After one of the above reactions occurs, you isolate \(2.27 \mathrm{~g}\) of copper. Which equation above describes the reaction that occurred? Support your answer.

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