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Chapter 13: Problem 44

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 \(0.500 \mathrm{~L}\) ?

Short Answer

Expert verified
The molarity for each solution is: (a) 175.2 M for Al2(SO4)3 (b) 0.1697 M for Mn(NO3)2路2H2O (c) 0.63 M for the diluted H2SO4

Step by step solution

01

Convert mass of solutes to moles

To find the moles of solute in each solution, we will use the given mass and the molar mass of the solute. The molar mass can be found by summing up the atomic masses of each element in the solute compound. (a) For Al2(SO4)3: Molar mass of Al2(SO4)3 = 2 脳 (Molar mass of Al) + 3 脳 (Molar mass of S) + 12 脳 (Molar mass of O) = 2 脳 26.98 + 3 脳 32.06 + 12 脳 16 = 53.96 + 96.18 + 192 = 342.14 g/mol 15.0 g of Al2(SO4)3 is equivalent to: \( \frac{15.0 \mathrm{~g}}{342.14 \mathrm{~g/mol}} = 0.0438 \mathrm{~moles} \) (b) For Mn(NO3)2路2H2O: Molar mass of Mn(NO3)2路2H2O = (Molar mass of Mn) + 2 脳 (Molar mass of N) + 6 脳 (Molar mass of O) + 4 脳 (Molar mass of H) = 54.94 + 2 脳 14.01 + 6 脳 16 + 4 脳 1.01 = 54.94 + 28.02 + 96 + 4.04 = 177.00 g/mol 5.25 g of Mn(NO3)2路2H2O is equivalent to: \( \frac{5.25 \mathrm{~g}}{177.00 \mathrm{~g/mol}} = 0.0297 \mathrm{~moles} \) (c) We are given the molarity and volume of a concentrated H2SO4 solution and need to find the molarity after dilution. First, we need to find the number of moles in the concentrated solution.
02

Calculate moles of solute in concentrated solution

Given molarity of H2SO4: 9.00 M Given initial volume of H2SO4: 35.0 mL = 0.035 L Using the formula for molarity: Moles of solute, \(n = Molarity \times Volume\) For H2SO4: \( n = 9.00 \mathrm{M} \times 0.035 \mathrm{L} = 0.315 \mathrm{~moles} \)
03

Calculate the molarity for each solution

(a) 0.0438 moles of Al2(SO4)3 in 0.250 mL solution: \( Molarity = \frac{0.0438 \mathrm{~moles}}{0.00025 \mathrm{L}} = 175.2 \mathrm{M} \) (b) 0.0297 moles of Mn(NO3)2路2H2O in 175 mL solution: \( Molarity = \frac{0.0297 \mathrm{~moles}}{0.175 \mathrm{L}} = 0.1697 \mathrm{M} \) (c) 0.315 moles of H2SO4 in 0.500 L: \( Molarity = \frac{0.315 \mathrm{~moles}}{0.500 \mathrm{L}} = 0.63 \mathrm{M} \) Thus, the molarity for each solution is: (a) 175.2 M for Al2(SO4)3 (b) 0.1697 M for Mn(NO3)2路2H2O (c) 0.63 M for the diluted H2SO4

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Key Concepts

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

Molarity Calculation
Molarity is a way to express the concentration of a solute in a solution. It indicates how many moles of a solute are present in one liter of solution. The formula to calculate molarity is:
  • \[ Molarity (M) = \frac{n}{V} \]
where:
  • \(n\) is the number of moles of solute
  • \(V\) is the volume of solution in liters
To calculate molarity, you first need the amount of solute in moles. You can find this by dividing the mass of the solute (in grams) by its molar mass (in \(g/mol\)).
For instance, if you have 15.0 grams of \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\), you calculate the moles of \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\) by dividing 15.0 by its molar mass 342.14 \(\mathrm{g/mol}\), resulting in 0.0438 moles.
The calculated moles is then divided by the solution's volume in liters to get the molarity.
Understanding how to compute molarity is essential for accurately determining the concentration of solutions in lab settings and theoretical calculations.
Chemical Compounds
Chemical compounds consist of elements that are combined in fixed ratios and structured by chemical bonds. Knowing the molecular composition of compounds is crucial for calculating their molar mass, which is necessary in various chemical calculations.
For example, \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\) is a compound made of aluminum, sulfur, and oxygen. By calculating the sum of the molar masses of these elements in their respective quantities, you ascertain the compound's molar mass.
\(\mathrm{Mn}(\mathrm{NO}_{3})_{2} \cdot 2 \mathrm{H}_{2}\mathrm{O}\) is a complex compound. It includes manganese, nitrogen, oxygen, and hydrogen. Calculating its molar mass involves evaluating the weights of each element, as well as accounting for the water molecules attached (2 \(\mathrm{H}_{2}\mathrm{O}\)).
  • For each component: sum up the atomic masses, multiply by occurrences within the compound, and add them together.
This detailed process allows you to convert the given mass of the compound into moles, an essential step in concentration calculations.
Dilution Process
Dilution is the process of reducing the concentration of a solute in a solution, typically by adding more solvent. The principle of dilution is that the amount of solute remains constant but spread across a larger volume of solvent. This can be expressed with the formula:
  • \[ C_1V_1 = C_2V_2 \]
Where:
  • \(C_1\) is the initial concentration
  • \(V_1\) is the initial volume
  • \(C_2\) is the final concentration
  • \(V_2\) is the final volume
For instance, starting with a 9.00 M concentration of \(\mathrm{H}_{2}\mathrm{SO}_{4}\), if 35.0 mL is diluted to 0.500 L, the final molarity can be calculated using the dilution equation. This results in finding that the new concentration is 0.63 M.
Practically, understanding dilution allows you to adjust the concentration for various experimental needs and is a vital technique for laboratory and industrial practices.

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Most popular questions from this chapter

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: CC(=O)NC(Cl)(Cl)Cl 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) Use the existence of hydrogen bonds between acetone and chloroform molecules to explain the deviation from ideal behavior. (c) 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. (d) Would you expect the same vaporpressure behavior for acetone and chloromethane \(\left(\mathrm{CH}_{3} \mathrm{Cl}\right)\) ? Explain.

An "emulsifying agent" is a compound that helps stabilize a hydrophobic colloid in a hydrophilic solvent (or a hydrophilic colloid in a hydrophobic solvent). Which of the following choices is the best emulsifying agent? (a) \(\mathrm{CH}_{3} \mathrm{COOH}\), (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\), (c) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{11} \mathrm{COOH}\), (d) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{11} \mathrm{COONa}\).

Calculate the molality of each of the following solutions: (a) \(8.66 \mathrm{~g}\) of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) dissolved in \(23.6 \mathrm{~g}\) of carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\), (b) \(4.80 \mathrm{~g}\) of \(\mathrm{NaCl}\) dissolved in \(0.350 \mathrm{~L}\) of water.

At ordinary body temperature \(\left(37^{\circ} \mathrm{C}\right)\), the solubility of \(\mathrm{N}_{2}\) in water at ordinary atmospheric pressure ( \(1.0 \mathrm{~atm})\) is \(0.015 \mathrm{~g} / \mathrm{L}\). Air is approximately \(78 \mathrm{~mol} \% \mathrm{~N}_{2}\). (a) Calculate the number of moles of \(\mathrm{N}_{2}\) dissolved per liter of blood, assuming blood is a simple aqueous solution. (b) At a depth of \(100 \mathrm{ft}\) in water, the external pressure is \(4.0 \mathrm{~atm}\). What is the solubility of \(\mathrm{N}_{2}\) from air in blood at this pressure? (c) If a scuba diver suddenly surfaces from this depth, how many milliliters of \(\mathrm{N}_{2}\) gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?

(a) A sample of hydrogen gas is generated in a closed container by reacting \(2.050 \mathrm{~g}\) of zinc metal with \(15.0 \mathrm{~mL}\) of \(1.00 \mathrm{M}\) sulfuric acid. Write the balanced equation for the reaction, and calculate the number of moles of hydrogen formed, assuming that the reaction is complete. (b) The volume over the solution in the container is \(122 \mathrm{~mL}\). Calculate the partial pressure of the hydrogen gas in this volume at \(25^{\circ} \mathrm{C}\), ignoring any solubility of the gas in the solution. (c) The Henry's law constant for hydrogen in water at \(25^{\circ} \mathrm{C}\) is \(7.8 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\)-atm. Estimate the number of moles of hydrogen gas that remain dissolved in the solution. What fraction of the gas molecules in the system is dissolved in the solution? Was it reasonable to ignore any dissolved hydrogen in part (b)? [13.111] The following table presents the solubilities of several gases in water at \(25^{\circ} \mathrm{C}\) under a total pressure of gas and water vapor of \(1 \mathrm{~atm}\). (a) What volume of \(\mathrm{CH}_{4}(\mathrm{~g})\) under standard conditions of temperature and pressure is contained in \(4.0 \mathrm{~L}\) of a saturated solution at \(25^{\circ} \mathrm{C}\) ? (b) Explain the variation in solubility among the hydrocarbons listed (the first three compounds), based on their molecular structures and intermolecular forces. (c) Compare the solubilities of \(\mathrm{O}_{2}, \mathrm{~N}_{2}\), and \(\mathrm{NO}\), and account for the variations based on molecular structures and intermolecular forces. (d) Account for the much larger values observed for \(\mathrm{H}_{2} \mathrm{~S}\) and \(\mathrm{SO}_{2}\) as compared with the other gases listed. (e) Find several pairs of substances with the same or nearly the same molecular masses (for example, \(\mathrm{C}_{2} \mathrm{H}_{4}\) and \(\mathrm{N}_{2}\) ), and use intermolecular interactions to explain the differences in their solubilities.
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