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Chapter 13: Problem 40

What is the molarity of each of the following solutions: (a) \(15.0 \mathrm{~g} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in \(0.350 \mathrm{~L}\) solution, (b) \(5.25 \mathrm{~g}\) \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) 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 molarities of the given solutions are (a) 0.125 M \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), (b) 0.124 M \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2} \cdot 2 \mathrm{H}_{2}\mathrm{O}\), and (c) 0.630 M \(\mathrm{H}_{2} \mathrm{SO}_{4}\).

Step by step solution

01

Calculate the moles of solute for each solution

We need to determine the moles of the solute by dividing the given mass by the molar mass of each solute. (a) For \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), we have 15.0 g of solute. The molar mass of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is: \[ (2 \times 26.98 + 3 \times (4 \times 16.00 + 1 \times 32.07)) \mathrm{g/mol} = 342.1 \;\mathrm{g/mol} \] (b) For \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2} \cdot 2 \mathrm{H}_{2}\mathrm{O}\), we have 5.25 g of solute. The molar mass is: \[ (54.94 + 2 \times (2 \times 16.00 + 3 \times 1.01 + 1 \times 14.01) + 2 \times 2 \times 1.01 + 2 \times 16.00) \mathrm{g/mol} = 241.05 \;\mathrm{g/mol} \] (c) The moles of solute needed for the third case can be calculated by multiplying the given concentration (9.00 M) with the given volume in liters (35.0 mL = 0.035 L).
02

Calculate moles of solute

Now, let's calculate the moles of solute for each solution: (a) Moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\): \[ \frac{15.0 \;\mathrm{g}}{342.1 \;\mathrm{g/mol}} = 0.0438 \;\mathrm{mol} \] (b) Moles of \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2} \cdot 2 \mathrm{H}_{2}\mathrm{O}\): \[ \frac{5.25 \;\mathrm{g}}{241.05 \;\mathrm{g/mol}} = 0.0218 \;\mathrm{mol} \] (c) Moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}\): \[ (9.00 \;\mathrm{M}) \times (0.035 \;\mathrm{L}) = 0.315 \;\mathrm{mol} \]
03

Determine the molarity for each solution

Finally, we'll divide the moles of solute by the volume of the solution in liters to determine the molarity: (a) Molarity of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\): \[ \frac{0.0438 \;\mathrm{mol}}{0.350 \;\mathrm{L}} = 0.125 \;\mathrm{M} \] (b) Molarity of \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2} \cdot 2 \mathrm{H}_{2}\mathrm{O}\): \[ \frac{0.0218 \;\mathrm{mol}}{0.175 \;\mathrm{L}} = 0.124 \;\mathrm{M} \] (c) Molarity of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) after dilution: \[ \frac{0.315 \;\mathrm{mol}}{0.500 \;\mathrm{L}} = 0.630 \;\mathrm{M} \] In summary, the molarities of the given solutions are (a) 0.125 M \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), (b) 0.124 M \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2} \cdot 2 \mathrm{H}_{2}\mathrm{O}\), and (c) 0.630 M \(\mathrm{H}_{2} \mathrm{SO}_{4}\).

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Key Concepts

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

Solution Preparation
Preparing a solution involves dissolving a specific amount of solute in a certain volume of solvent. This forms a solution of desired concentration. To make a solution:
  • Determine the amount of solute needed based on the desired molarity and volume of solution.
  • Use a balance to measure the exact mass of the solute.
  • Dissolve the solute in a part of the solvent first, ensuring it is completely dissolved.
  • Add more solvent until you reach the desired final volume.
This procedure ensures the solution has the correct concentration, calculated as moles of solute per liter of solution (molarity).
In our example, solutions contain different solutes like \({Al}_2{(SO_4)}_3\) or \({Mn}{(NO_3)}_2 \), and their preparation is crucial for accurate molarity determination.
Moles of Solute
Calculating the moles of solute is a critical step in solution preparation. Moles quantify the amount of solute molecules present.
  • Start by measuring the mass of your solute.
  • Determine the molar mass of the solute using its chemical formula.
  • Convert the mass to moles using the formula:
    \[\text{moles} = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}}\]
For example, if you have 15g of \({Al}_2{(SO_4)}_3\), use its molar mass (342.1 g/mol) to find the moles: \(15 \div 342.1 = 0.0438 \text{ mol}\).
Understanding moles is key for finding the concentration of a solution.
Dilution
Dilution involves adding more solvent to a solution to decrease its concentration. This technique is useful for achieving desired molarity from a more concentrated solution. Considerations for dilution:
  • Know the initial concentration (molarity) and volume of the solution.
  • Decide on the final volume needed.
  • Apply the dilution formula:
    \[M_1 \times V_1 = M_2 \times V_2\]
In the formula, \(M_1\) and \(V_1\) represent the initial molarity and volume, while \(M_2\) and \(V_2\) represent the final molarity and volume.
For example, if you dilute 35 mL of 9M \(H_2SO_4\) to 500 mL, the new molarity is found using \(0.035 \times 9 = 0.5 \times M_2\), resulting in \(M_2 = 0.63M\).
Dilution is effective for adjusting solution strength.
Molar Mass Calculation
Molar mass is the sum of the masses of all atoms in a molecule. It's an essential component in converting between grams and moles.
  • Identify each element in the molecule and its number of atoms.
  • Multiply the atomic mass of each element (found on the periodic table) by the number of atoms.
  • Add these values to find the total molar mass.
For example, to find the molar mass of \(\text{Mn}{(NO}_3)_2 \cdot 2 \text{H}_2\text{O}\):
Add masses:
\(\text{Mn}=54.94\), \(\text{N}=14.01\), \(\text{O}=16.00\), \(\text{H}=1.01\)
Results in \(241.05 \, \text{g/mol}\).
Calculating molar mass supports accurate conversion to moles, crucial for solution preparation and concentration calculations.

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Most popular questions from this chapter

Consider two solutions, one formed by adding \(10 \mathrm{~g}\) of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) to \(1 \mathrm{~L}\) of water and the other formed by adding \(10 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) to \(1 \mathrm{~L}\) of water. Are the vapor pressures over the two solutions the same? Why or why not?

(a) What is an ideal solution? (b) The vapor pressure of pure water at \(60^{\circ} \mathrm{C}\) is 149 torr. The vapor pressure of water over a solution at \(60^{\circ} \mathrm{C}\) containing equal numbers of moles of water and ethylene glycol (a nonvolatile solute) is 67 torr. Is the solution ideal according to Raoult's law? Explain.

In general, the attractive intermolecular forces between solvent and solute particles must be comparable or greater than solute-solute interactions for significant solubility to occur. Explain this statement in terms of the overall energetics of solution formation.

(a) Many proteins that remain homogeneously distributed in water have molecular masses in the range of 30,000 amu and larger. In what sense is it appropriate to consider such suspensions to be colloids rather than solutions? Explain. (b) What general name is given to a colloidal dispersion of oneliquid in another? What is an emulsifying agent?

Describe how you would prepare each of the following aqueous solutions: (a) \(1.50 \mathrm{~L}\) of \(0.110 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) solution, starting with solid \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} ;\) (b) \(120 \mathrm{~g}\) of a solution that is \(0.65 \mathrm{~m}\) in \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), starting with the solid solute; (c) \(1.20 \mathrm{~L}\) of a solution that is \(15.0 \% \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}\) by mass (the density of the solution is \(1.16 \mathrm{~g} / \mathrm{mL}\) ), starting with solid solute; (d) a \(0.50 \mathrm{M}\) solution of \(\mathrm{HCl}\) that would just neutralize \(5.5 \mathrm{~g}\) of \(\mathrm{Ba}(\mathrm{OH})_{2}\) starting with \(6.0\) M HCl.
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