Problem 43 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

The molarity of the solutions is as follows: (a) \(1.46 \times 10^{-2}\) M (b) 1.118 M (c) 0.35 M

Step 1: Part (a): Calculate the moles of Mg(NO3)2

To determine the moles of Mg(NO3)2, we will first need to find its molar mass. The molar mass of Mg(NO3)2 can be calculated by adding the molar masses of each element in the compound, which are: - Magnesium (Mg): 24.31 g/mol - Nitrogen (N): 14.01 g/mol - Oxygen (O): 16.00 g/mol Molar mass of Mg(NO3)2 = 1(24.31) + 2(14.01 + 3(16.00)) = 24.31 + 2(62.03) = 148.37 g/mol Now, use the mass of Mg(NO3)2 and its molar mass to determine the moles of Mg(NO3)2: moles of Mg(NO3)2 = 0.540 g / 148.37 g/mol = \(3.64 \times 10^{-3}\) mol

Step 2: Part (a): Convert the volume of the solution to liters

The given volume of the solution is 250.0 mL. To convert it to liters, divide the volume by 1000: volume in liters = 250.0 mL / 1000 = 0.250 L

Step 3: Part (a): Calculate the molarity of the solution

Now, we can use the moles of Mg(NO3)2 and the volume in liters to calculate the molarity: M = moles of solute/volume of solution in liters = \(3.64 \times 10^{-3}\) mol / 0.250 L = \(1.46 \times 10^{-2}\) M

Step 4: Part (b): Calculate the moles of LiClO4 . 3H2O

To determine the moles of LiClO4 . 3H2O, we will first need to find its molar mass. The molar mass of LiClO4 . 3H2O can be calculated by adding the molar masses of each element in the compound, which are: - Lithium (Li): 6.94 g/mol - Chlorine (Cl): 35.45 g/mol - Oxygen (O): 16.00 g/mol - Hydrogen (H): 1.01 g/mol Molar mass of LiClO4 . 3H2O = 1(6.94) + 1(35.45) + 4(16.00) + 6(1.01) + 3(16.00) = 6.94 + 35.45 + 64.00 + 6.06 + 48.00 = 160.45 g/mol Now, use the mass of LiClO4 . 3H2O and its molar mass to determine the moles of LiClO4 . 3H2O: moles of LiClO4 . 3H2O = 22.4 g / 160.45 g/mol = 0.1397 mol

Step 5: Part (b): Convert the volume of the solution to liters

The given volume of the solution is 125 mL. To convert it to liters, divide the volume by 1000: volume in liters = 125 mL / 1000 = 0.125 L

Step 6: Part (b): Calculate the molarity of the solution

Now, we can use the moles of LiClO4 . 3H2O and the volume in liters to calculate the molarity: M = moles of solute/volume of solution in liters = 0.1397 mol / 0.125 L = 1.118 M

Step 7: Part (c): Dilution equation

In this part, we have a concentrated solution of HNO3 that is diluted. The dilution equation can be used to determine the concentration of the diluted solution: Mi x Vi = Mf x Vf where Mi is the initial molarity, Vi is the initial volume, Mf is the final molarity, and Vf is the final volume.

Step 8: Part (c): Calculate the molarity of the diluted solution

In this case, Mi = 3.50 M, Vi = 25.0 mL, and Vf = 250 mL. We need to determine Mf: Insert the known values into the equation and convert mL to L: (3.50 M)(0.025 L) = Mf(0.250 L) Now, solve for Mf: Mf = (3.50 M)(0.025 L)/(0.250 L) = 0.35 M The molarity of the diluted HNO3 solution is 0.35 M

Key Concepts

Aqueous Solutions

An aqueous solution is a mixture where water is the solvent, and another substance, called the solute, is dissolved in it. Molarity calculations focus on determining how much of the solute is present in a certain volume of the solution.
This is crucial when you need to know the strength or concentration of the solution.

For example, in the given exercise, we are dealing with solutions like magnesium nitrate or lithium perchlorate trihydrate dissolved in water. The main goal is to calculate their molarity, which tells us how many moles of solute are present per liter of solution.

• The benefit of aqueous solutions is that they allow chemical reactions to occur efficiently as ions or molecules are free to move and react.
• Knowing molarity helps predict how a solution will behave when mixed with others.

Understanding the concept of an aqueous solution is foundational to mastering various chemistry problems, including those involving molarity calculations.

Dilution Equation

Dilution plays a vital role when preparing solutions of desired concentrations from more concentrated ones. The dilution equation is expressed as \( M_i \times V_i = M_f \times V_f \). Here, \( M_i \) is the initial molarity, \( V_i \) the initial volume, \( M_f \) the final molarity, and \( V_f \) the final volume of the solution.
This equation ensures that the number of solute moles remains constant before and after dilution.

Let's consider the example from the problem where nitric acid (HNO3) is diluted. The initial conditions of the concentrated solution allow us to apply the equation and find the diluted molarity.

• It's crucial to match units properly, typically converting volumes into liters to maintain consistency.
• Understanding this equation helps manipulate solutions to achieve precise concentrations needed for experiments or applications.

Proper use of the dilution equation is significant for lab work and real-world applications, ensuring solutions are safe and effective.

Molar Mass

Molar mass is an essential concept in chemistry when dealing with molarity calculations. It gives us the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Calculating the molar mass of compounds is crucial for finding the number of moles present in a given mass of the substance.

In exercises like the one given, knowing the molar mass of compounds such as magnesium nitrate or lithium perchlorate trihydrate allows for determining their moles using the formula:\[\text{Moles} = \frac{\text{Mass of substance}}{\text{Molar mass}}\]
It's important to add the molar masses of all elements in a compound accurately.

• This knowledge is critical for calculating the molarity of a solution since it relates the mass of a solute to the solution volume.
• Molar masses are readily available in periodic tables, making it easy to handle such calculations.

Grasping how to determine and use molar mass is fundamental for chemistry problems involving reactions, stoichiometry, and preparing solutions.