Problem 44 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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 × (Molar mass of Al) + 3 × (Molar mass of S) + 12 × (Molar mass of O) = 2 × 26.98 + 3 × 32.06 + 12 × 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)2·2H2O: Molar mass of Mn(NO3)2·2H2O = (Molar mass of Mn) + 2 × (Molar mass of N) + 6 × (Molar mass of O) + 4 × (Molar mass of H) = 54.94 + 2 × 14.01 + 6 × 16 + 4 × 1.01 = 54.94 + 28.02 + 96 + 4.04 = 177.00 g/mol 5.25 g of Mn(NO3)2·2H2O 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.

Step 2: 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: \( n = 9.00 \mathrm{M} \times 0.035 \mathrm{L} = 0.315 \mathrm{~moles} \)

Step 3: 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)2·2H2O 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)2·2H2O (c) 0.63 M for the diluted H2SO4

Key Concepts

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.