Question

Calculate the molarity of the following aqueous solutions: (a) \(0.540 \mathrm{~g} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) in \(250.0 \mathrm{~mL}\) of solution, (b) \(22.4 \mathrm{~g} \mathrm{LiClO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}\) in \(125 \mathrm{~mL}\) of solution, (c) \(25.0 \mathrm{~mL}\) of \(3.50 \mathrm{M} \mathrm{HNO}_{3}\) diluted to \(0.250 \mathrm{~L}\).

Short answer

The molarity of the solutions are: (a) approximately 0.0146 M for Mg(NO3)2, (b) approximately 0.564 M for LiClO4·3H2O, and (c) approximately 0.350 M for the diluted HNO3 solution.

Step-by-step solution

(a) Molecular weight of Mg(NO3)2

(Assuming you meant, 0.540 g Mg(NO3)2) First, we need to determine the molecular weight of magnesium nitrate, Mg(NO3)2. The molecular weight of Mg(NO3)2 is the sum of the atomic weights of its elements: \(Molecular~ weight~ (Mg(NO_{3})_{2}) = 1 \times atomic~ weight~ (Mg) + 2 \times [ 1 \times atomic~ weight~ (N) + 3 \times atomic~ weight~ (O) ]\) Using atomic weights from the periodic table: \(Molecular~ weight~ (Mg(NO_{3})_{2}) = 1 \times 24.31 + 2 \times [1 \times 14.01 + 3 \times 16.00]\) \(Molecular~ weight~ (Mg(NO_{3})_{2}) = 24.31 + 2 \times 62.01\) \(Molecular~ weight~ (Mg(NO_{3})_{2}) = 148.33~ g/mol\)

(a) Moles of solute

To calculate the moles of magnesium nitrate, divide the mass of the solute by its molecular weight: \(\mathrm{Moles_Mg(NO_{3})_{2}} = \frac{0.540~g}{148.33~g/mol} = 0.00364~mol\)

(a) Molarity calculation

To calculate the molarity of the solution, divide the moles of solute by the volume of the solution in liters: \(\mathrm{Molarity}=\frac{0.00364~mol}{0.2500~L} = 0.0146~M\) The molarity of the Mg(NO3)2 solution is approximately 0.0146 M.

(b) Molecular weight of LiClO4·3H2O

First, determine the molecular weight of lithium perchlorate trihydrate (LiClO4·3H2O). The molecular weight is the sum of the atomic weights of its elements: \(Molecular~ weight~ (LiClO_4\cdot3H_2O) = 1 \times atomic~ weight~ (Li) + 1 \times atomic~ weight~ (Cl) + 4 \times atomic~ weight~ (O) + 3 \times [2 \times atomic~ weight~ (H) + 1 \times atomic~ weight~ (O) ]\) Using atomic weights from the periodic table: \(Molecular~ weight~ (LiClO_4\cdot3H_2O) = 1 \times 6.94 + 1 \times 35.45 + 4 \times 16.00 + 3 \times [2 \times 1.008 + 1 \times 16.00]\) \(Molecular~ weight~ (LiClO_4\cdot3H_2O) = 318.16~ g/mol\)

(b) Moles of solute

To calculate the moles of lithium perchlorate trihydrate, divide the mass of the solute by its molecular weight: \(\mathrm{Moles_{LiClO_{4}\cdot3H_{2}O}}=\frac{22.4~g}{318.16~g/mol} = 0.0705~mol\)

(b) Molarity calculation

To calculate the molarity of the solution, divide the moles of solute by the volume of the solution in liters: \(\mathrm{Molarity} =\frac{0.0705~mol}{0.125~L} = 0.564~M\) The molarity of the LiClO4·3H2O solution is approximately 0.564 M.

(c) Molarity calculation

In this case, we already have a concentration for the initial HNO3 solution (3.50 M) and the diluted volume of the final solution (0.250 L). To calculate the molarity of the diluted solution, we can use the dilution formula: \(M_1V_1 = M_2V_2\) Where \(M_1\) and \(V_1\) are the initial molarity and volume, and \(M_2\) and \(V_2\) are the final molarity and volume. Rearrange the formula to find the final molarity, \(M_2\): \(\mathrm{M_{2}}=\frac{\mathrm{M_{1}} \cdot \mathrm{V_{1}}}{\mathrm{V_{2}}}\) Given the initial volume (25.0 mL) and the final volume (0.250 L), we can calculate the final molarity: \(\mathrm{M_{2}} = \frac{3.50~M \cdot 0.0250~L}{0.250~L} = 0.350~M\) The molarity of the diluted HNO3 solution is approximately 0.350 M.

Key concepts

Understanding Molar Concentration

Molar concentration, often referred to as molarity, is a measure used in chemistry to express the concentration of a solute in a solution. It is defined as the number of moles of a substance divided by the volume of the solution in liters. Expressed in one equation, molarity (M) is calculated as:

\[ Molarity~(M) = \frac{\text{moles of solute}}{\text{liters of solution}} \]

In the given exercise, molarity calculations are essential for determining the strength of the aqueous solutions. To successfully calculate molarity, it is imperative to first know the moles of the substance, which brings us to the concept of molecular weight.

The Crucial Role of Molecular Weight in Calculations

The term 'molecular weight' refers to the mass of one mole of a substance, often expressed in grams per mole (g/mol). To ascertain the molecular weight, one must take account of the individual atomic masses of each element that constitutes a molecule. For example:

\[ Molecular~weight~(Compound) = \sum \text{(number of atoms of each element)} \times \text{(atomic weight of each element)} \]

What is Atomic Weight and How is it Used?

Atomic weight is the average mass of atoms of an element, measured in atomic mass units (amu). It considers the various isotopes of an element and their natural abundance. When it comes to carrying out molarity calculations, knowing the molecular weight allows us to convert the given mass of a solute into moles—a necessary step for arriving at molarity.

Utilizing the Dilution Formula

Dilution is a process by which the concentration of a solution is reduced. The dilution formula reflects the relationship between the concentrations and volumes before and after the dilution process:

\[ M_1V_1 = M_2V_2 \]

Where \(M_1\) is the initial molarity, \(V_1\) is the initial volume, \(M_2\) is the final molarity, and \(V_2\) is the final volume. By rearranging the formula to solve for the unknown concentration or volume, it becomes an indispensable tool in the laboratory. In our exercise scenario, the formula applied to the dilution of an HNO3 solution results in understanding how molarity changes with the introduction of additional solvent.

To make this concept clear to students, one could suggest employing real-life examples such as how a concentrated juice is diluted before consumption, which illustrates the concept of dilutions in a more relatable context.