Problem 43 - Chemistry The Central Science

Question

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

Step-by-Step Solution

Answer

The molarities of the given aqueous solutions are approximately: (a) 0.0146 M Mg(NO3)2 (b) 0.682 M LiClO4.3H2O (c) 0.350 M HNO3

Step 1: (a) Calculate the Molar Mass of Mg(NO3)2

To find the molarity of the Mg(NO3)2 solution, first calculate the molar mass of Mg(NO3)2. The molar mass of Mg(NO3)2 is obtained by adding the atomic masses of its constituent elements: Mg(NO3)2 = Mg + 2 x (N + 3 x O) = 24.31 g/mol (Mg) + 2 x [14.01 g/mol (N) + 3 x 16.00 g/mol (O)] = 24.31 g/mol + 2 x (14.01 g/mol + 48.00 g/mol) = 24.31 g/mol + 2 x 62.01 g/mol = 148.33 g/mol

Step 2: (a) Calculate moles of Mg(NO3)2

Now you can calculate the moles of Mg(NO3)2 by dividing the mass given (0.540 g) by its molar mass: Moles of Mg(NO3)2 = (0.540 g) / (148.33 g/mol) ≈ 0.00364 mol

Step 3: (a) Convert mL to L

Next, convert the volume of the solution from mL to L: 250.0 mL = 250.0 / 1000 = 0.250 L

Step 4: (a) Calculate Molarity of Mg(NO3)2 Solution

Using the molarity formula (M = moles of solute / liters of solution), you can find the molarity of this solution: Molarity of Mg(NO3)2 = (0.00364 mol) / (0.250 L) ≈ 0.0146 M

Step 5: (b) Calculate the Molar Mass of LiClO4.3H2O

First, find the molar mass of LiClO4.3H2O by adding the atomic masses of its constituent elements: LiClO4.3H2O = Li + Cl + 4 x O + 3 x (2 x H2O) = 6.94 g/mol (Li) + 35.45 g/mol (Cl) + 4 x 16.00 g/mol (O) + 3 x (2 x 1.01 g/mol) (H2O) = 6.94 g/mol + 35.45 g/mol + 64.00 g/mol + 3 x 36.03 g/mol = 262.94 g/mol

Step 6: (b) Calculate moles of LiClO4.3H2O

Now, calculate the moles of LiClO4.3H2O by dividing the given mass of LiClO4.3H2O (22.4 g) by its molar mass: Moles of LiClO4.3H2O = (22.4 g) / (262.94 g/mol) = 0.0852 mol

Step 7: (b) Convert mL to L

Next, convert the volume of the LiClO4.3H2O solution from mL to L: 125 mL = 125 / 1000 = 0.125 L

Step 8: (b) Calculate Molarity of LiClO4.3H2O Solution

Now, use the molarity formula to find the molarity of the LiClO4.3H2O solution: Molarity of LiClO4.3H2O = (0.0852 mol) / (0.125 L) ≈ 0.682 M

Step 9: (c) Find Initial Moles of HNO3

For this solution, you are given the initial concentration (3.50 M) and volume (25.0 mL) of HNO3. First, convert the volume to liters: 25.0 mL = 25.0 / 1000 = 0.025 L Now find the initial moles of HNO3: Initial moles of HNO3 = Initial concentration x Initial volume = (3.50 mol/L) x (0.025 L) ≈ 0.0875 mol

Step 10: (c) Calculate Molarity of Diluted HNO3 Solution

Now, calculate the molarity of the diluted solution using the moles of HNO3 and the final volume (0.250 L): Molarity of diluted HNO3 solution = (0.0875 mol) / (0.250 L) ≈ 0.350 M The molarities of the solutions are approximately as follows: (a) 0.0146 M Mg(NO3)2 (b) 0.682 M LiClO4.3H2O (c) 0.350 M HNO3

Key Concepts

Molar Mass Determination

Understanding molar mass is essential for many chemistry calculations. It represents the mass of one mole of a substance, expressed in grams per mole (g/mol). To calculate the molar mass of a compound, like Mg(NO₃)₂ or LiClO₄·3H₂O from our exercise, you need to sum the atomic masses of each element present in the compound, taking into account their respective quantity in the molecule.

For instance, Mg(NO₃)₂ includes one magnesium atom, two nitrogen atoms, and six oxygen atoms. By adding the atomic masses of these elements, using values from the periodic table, you determine the molar mass. This step is fundamental, as the molar mass is then used to convert the mass of the compound into moles, a key step in calculating molarity.

Moles and Molar Concentration

The concept of moles is at the heart of chemistry. It's a measure of quantity that allows chemists to count particles like atoms and molecules. One mole contains Avogadro's number of particles, which is approximately 6.022 × 10²³. To move from grams to moles, you divide the mass of a substance by its molar mass.

Molar concentration, often simply called molarity and expressed as 'M', is the number of moles of solute (the substance being dissolved) per liter of solution. The formula to calculate molarity is M = moles of solute / liters of solution. In our exercise, once we calculate the number of moles of our substance, we use this formula to find the molarity, allowing us to understand the concentration of the solution.

Solution Dilution

Dilution involves adding more solvent to a solution to decrease its concentration. The amount of solute remains constant; only the total volume changes. It's crucial to understand that the concentration of a solution is inversely related to its volume when diluting.

Using the formula M₁V₁ = M₂V₂, where M₁ is the initial molarity, V₁ is the initial volume, M₂ is the final molarity, and V₂ is the final volume, you can calculate the molarity of a diluted solution. In the given exercise, the solution of HNO₃ was diluted, and by applying this concept, we can determine its new molarity.

Volume to Liters Conversion

In chemistry experiments, it's often necessary to convert volume measurements to liters, as molarity calculations are based on volume in liters. In everyday life, we encounter various volume measurements, such as milliliters, cups, or gallons. However, since one liter is defined as 1000 milliliters, converting milliliters to liters simply involves dividing the volume by 1000.

In our textbook problem, for example, converting 250.0 mL to liters involves dividing by 1000, yielding 0.250 liters. This conversion is vital for the accurate determination of molarity and is a fundamental part of solution-based calculations in chemistry.