Problem 68 - Chemistry The Central Science

Question

(a) How many grams of solute are present in \(15.0 \mathrm{~mL}\) of \(0.736 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} ?\) (b) If \(14.00 \mathrm{~g}\) of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) is dissolved in enough water to form \(250 \mathrm{~mL}\) of solution, what is the molarity of the solution? (c) How many milliliters of \(0.0455 \mathrm{M} \mathrm{CuSO}_{4}\) contain \(3.65 \mathrm{~g}\) of solute?

Step-by-Step Solution

Answer

(a) There are \(3.25 \mathrm{~g}\) of solute present in \(15.0 \mathrm{~mL}\) of \(0.736 \mathrm{M K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) solution. (b) The molarity of the solution formed by dissolving \(14.00 \mathrm{~g}\) of \(\left(\mathrm{NH}_{4}\right)_{2}\mathrm{SO}_{4}\) in \(250 \mathrm{~mL}\) of water is \(0.424 \mathrm{M}\). (c) \(503.1 \mathrm{~mL}\) of \(0.0455 \mathrm{M CuSO}_{4}\) solution contains \(3.65 \mathrm{~g}\) of solute.

Step 1: Identify the given information

The given information is: - Volume of the solution: \(15.0 \mathrm{~mL}\) - Molarity of the solution: \(0.736 \mathrm{M}\) - Substance: \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\)

Step 2: Determine the molar mass of solute

Use the periodic table to find the molar mass of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\): Molar mass of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} = 2\times\mathrm{M}_\mathrm{K} + 2\times\mathrm{M}_\mathrm{Cr} + 7\times\mathrm{M}_\mathrm{O} = 2(39.10 \mathrm{~g/mol}) + 2(51.996\mathrm{~g/mol}) + 7(16.00\mathrm{~g/mol}) = 294.2 \mathrm{~g/mol}\)

Step 3: Calculate moles of solute

Use the molarity formula to find moles of solute. Remember that molarity (\(M\)) is the moles of solute per liter of solution: \(0.736 \mathrm{~M} = \frac{\text{moles of solute}}{0.015 \mathrm{~L}}\) \(\text{moles of solute} = 0.736 \mathrm{~M} \times 0.015 \mathrm{~L} = 0.01104 \mathrm{~mol}\)

Step 4: Calculate grams of solute

Multiply the moles of solute by the molar mass of the solute to find the grams of solute present: \(\text{grams of solute} = 0.01104 \mathrm{~mol} \times 294.2 \mathrm{~g/mol} = 3.25 \mathrm{~g}\) (b) Determining the molarity of a solution

Step 1: Identify the given information

The given information is: - Mass of solute: \(14.00 \mathrm{~g}\) - Volume of the solution: \(250 \mathrm{~mL}\) - Substance: \(\left(\mathrm{NH}_{4}\right)_{2}\mathrm{SO}_{4} \)

Step 2: Determine the molar mass of solute

Use the periodic table to find the molar mass of \(\left(\mathrm{NH}_{4}\right)_{2}\mathrm{SO}_{4} \): Molar mass of \(\left(\mathrm{NH}_{4}\right)_{2}\mathrm{SO}_{4} = 2\times(14.01\mathrm{~g/mol} + 4\times1.008\mathrm{~g/mol}) + 32.07\mathrm{~g/mol} + 4\times16.00\mathrm{~g/mol} = 132.14 \mathrm{~g/mol} \)

Step 3: Calculate moles of solute

Divide the mass of solute by the molar mass of the solute to find the moles of solute present: \(\text{moles of solute} = \frac{14.00 \mathrm{~g}}{132.14 \mathrm{~g/mol}} = 0.10593 \mathrm{~mol}\)

Step 4: Calculate molarity of the solution

Use the molarity formula with the moles of solute and volume of solution: \(M = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{0.10593 \mathrm{~mol}}{0.250 \mathrm{~L}} = 0.424 \mathrm{M}\) (c) Finding the volume of a solution containing a certain mass

Step 1: Identify the given information

The given information is: - Mass of solute: \(3.65 \mathrm{~g}\) - Molarity of the solution: \(0.0455 \mathrm{M}\) - Substance: \(\mathrm{CuSO}_{4}\)

Step 2: Determine the molar mass of solute

Use the periodic table to find the molar mass of \(\mathrm{CuSO}_{4}\): Molar mass of \(\mathrm{CuSO}_{4} = 63.546\mathrm{~g/mol} + 32.07\mathrm{~g/mol} + 4\times16.00\mathrm{~g/mol} = 159.62 \mathrm{~g/mol}\)

Step 3: Calculate moles of solute

Divide the mass of solute by the molar mass of the solute to find the moles of solute present: \(\text{moles of solute} = \frac{3.65 \mathrm{~g}}{159.62 \mathrm{~g/mol}} = 0.02288 \mathrm{~mol}\)

Step 4: Calculate the volume of the solution

Use the molarity formula with the moles of solute and the molarity of the solution: \(\text{volume of solution} = \frac{\text{moles of solute}}{\text{Molarity}} = \frac{0.02288 \mathrm{~mol}}{0.0455 \mathrm{M}} = 0.5031 \mathrm{L}\)

Step 5: Convert volume to milliliters

Multiply the volume in liters by 1000 to convert it to milliliters: \(\text{volume of solution} = 0.5031 \mathrm{L} \times 1000 = 503.1 \mathrm{~mL}\)

Key Concepts

Solubility

Solubility is a fundamental concept in chemistry that refers to the ability of a solute to dissolve in a solvent to form a homogeneous solution. The solubility of a substance depends on both physical and chemical properties, including temperature and pressure.
It's important to note that solubility can be quantitatively expressed as the maximum amount of solute that can dissolve in a given volume of solvent at a specific temperature. This is often described in terms of grams per liter (g/L). For instance:

• When a solute like potassium dichromate ( K_2Cr_2O_7), is dissolved in water, how much dissolves depends on its intrinsic solubility.
• Different substances have different solubility levels.

Moreover, solubility can be influenced by the presence of other substances in the solution, like salts or ions, which can either increase or decrease the solubility of the solute. Understanding solubility helps in calculating how much of a solute is needed to reach saturation, or how to prepare specific solution concentrations for reactions and processes.

Moles

The concept of moles is central in chemistry for quantifying the amount of a substance. A mole represents Avogadro's number, approximately 6.022 × 10^23, of particles, which can be atoms, molecules, or ions depending on the substance.
Moles are used for expressing amounts of reactants and products in chemical reactions, which makes it easier for chemists to relate masses to moleculeslearn more:

• Calculations start with the mass of a substance, which is converted to moles using its molar mass.
• The mole concept ensures a coherent way to figure out how substances will interact in a solution, such as our case with copper(II) sulfate (CuSO_4).
• It is also a standard that helps compare how much of one substance reacts with another properly.

While moles measure a quantity in terms of particles, they allow chemical equations to be balanced and provide a concrete way to study stoichiometry in reactions.

Molar Mass

Molar mass bridges the gap between the atomic world and the macroscopic world. It is defined as the mass of one mole of a given substance and is expressed in grams per mole (g/mol). It is calculated by summing up the atomic masses of each element present in a compound, with each element multiplied by its respective quantity in the formula:

• Molar mass of a compound allows conversion from the mass of a substance to the amount in moles, an essential step in stoichiometric calculations.
• For our examples, this concept was used to determine the grams of specific solutes like K_2Cr_2O_7 and (NH_4)_2SO_4, which have molar masses of 294.2 g/mol and 132.14 g/mol, respectively.
• By understanding molar mass, chemists can predict the outcomes of chemical reactions by calculating how much reactant is needed to produce a certain amount of product.

This fundamental property is crucial for performing calculations involving gases, liquids, or solids, especially in preparing solutions.

Volume Conversion

Volume conversion is an indispensable calculation step when dealing with liquid solutions. It allows chemists to express volumes in different units, most commonly liters (L) and milliliters (mL)
Efficient conversion aids in the precise measurements required for chemical reactions.

• In chemical solutions, often you need to convert between milliliters and liters since molarity is generally expressed in terms of liters.
• For instance, in our exercise, converting the solution's volume from mL to L allows the correct use of the molarity formula.
• Knowing that 1 L = 1000 mL, students can easily convert these units back and forth as required.

Accurate volume conversion is crucial to ensure that solution concentrations are prepared correctly, which impacts both the effectiveness and the outcome of chemical experiments or productions.