Question

Calculate the molarity of each of these solutions. a. A \(5.623-\mathrm{g}\) sample of \(\mathrm{NaHCO}_{3}\) is dissolved in enough water to make \(250.0 \mathrm{~mL}\) of solution. b. A \(184.6-\mathrm{mg}\) sample of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is dissolved in enough water to make \(500.0 \mathrm{~mL}\) of solution. c. A 0.1025-g sample of copper metal is dissolved in \(35 \mathrm{~mL}\) of concentrated \(\mathrm{HNO}_{3}\) to form \(\mathrm{Cu}^{2+}\) ions and then water is added to make a total volume of \(200.0 \mathrm{~mL}\). (Calculate the molarity of \(\mathrm{Cu}^{2+} .\) )

Short answer

The molarity of each solution is as follows: a. \(0.2676\,M\) NaHCO3 b. \(0.001256\,M\) K2Cr2O7 c. \(0.008065\,M\) Cu2+

Step-by-step solution

Calculate the moles of NaHCO3

The molar mass of NaHCO3 can be calculated as follows: Molar mass of Na = 22.99 g/mol; Molar mass of H = 1.01 g/mol; Molar mass of C = 12.01 g/mol; Molar mass of O = 16.00 g/mol. Molar mass of NaHCO3 = 22.99 + 1.01 + 12.01 + (3 × 16.00) = 84.01 g/mol. Now, to find the moles in 5.623 g of NaHCO3, we use the formula: moles = mass / molar mass = \( \frac{5.623}{84.01} \) = 0.0669 mol.

Convert the volume of the solution to liters

We are given that the volume of the solution is 250.0 mL. To convert this to liters, we divide by 1000: Volume in liters = 250.0/1000 = 0.250 L.

Calculate the molarity of the solution

Molarity (M) is calculated by dividing the moles of solute by the volume of the solution in liters: M = \( \frac{0.0669}{0.250} \) = 0.2676 M. The molarity of NaHCO3 in the solution is 0.2676 M. ## Solution for b. A 184.6-mg sample of K2Cr2O7 ##

Calculate the moles of K2Cr2O7

The molar mass of K2Cr2O7 can be calculated as follows: Molar mass of K = 39.10 g/mol; Molar mass of Cr = 51.99 g/mol; Molar mass of O = 16.00 g/mol. Molar mass of K2Cr2O7 = (2 × 39.10) + (2 × 51.99) + (7 × 16.00) = 294.18 g/mol. Now, to find the moles in 184.6 mg (0.1846 g) of K2Cr2O7, we use the formula: moles = mass / molar mass = \( \frac{0.1846}{294.18} \) = 0.000628 mol.

Convert the volume of the solution to liters

We are given that the volume of the solution is 500.0 mL. To convert this to liters, we divide by 1000: Volume in liters = 500.0/1000 = 0.500 L.

Calculate the molarity of the solution

Molarity (M) is calculated by dividing the moles of solute by the volume of the solution in liters: M = \( \frac{0.000628}{0.500} \) = 0.001256 M. The molarity of K2Cr2O7 in the solution is 0.001256 M. ## Solution for c. A 0.1025-g sample of copper metal with Cu2+ ions ##

Calculate the moles of Cu2+

The molar mass of Cu is 63.55 g/mol. To find the moles in 0.1025 g of Cu, we use the formula: moles = mass / molar mass = \( \frac{0.1025}{63.55} \) = 0.001613 mol.

Convert the volume of the solution to liters

We are given that the final volume of the solution is 200.0 mL. To convert this to liters, we divide by 1000: Volume in liters = 200.0/1000 = 0.200 L.

Calculate the molarity of the solution

Molarity (M) is calculated by dividing the moles of solute by the volume of the solution in liters: M = \( \frac{0.001613}{0.200} \) = 0.008065 M. The molarity of Cu2+ in the solution is 0.008065 M.

Key concepts

Concentration of Solution

Understanding the concentration of a solution is crucial when working in chemistry. It allows scientists and students alike to describe how much solute is present in a given amount of solvent. Molarity, often represented as M, is the measure of concentration used in the above exercises. It is defined as the number of moles of solute per liter of solution.

The molarity calculation is straightforward: Divide the number of moles of solute by the volume of the solution in liters. For instance, in solution a for NaHCO₃, after calculating the number of moles, we divide it by the total volume in liters, yielding the molarity of the solution. This concept is pivotal not just for laboratory tasks but also for understanding reactions in a broader context, such as in environmental chemistry or medical applications.

Moles and Molar Mass

In the midst of these calculations, the terms moles and molar mass are indispensable. A mole is a unit that represents a specific number of particles, approximately 6.022 x 10²³, known as Avogadro's number. The molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol).

For precise calculations like those demonstrated in the problem, you first determine the molar mass by adding the atomic masses of all atoms in a compound. Next, you calculate the moles of the substance by dividing its mass by its molar mass. These steps underpin many processes in chemistry, from reaction stoichiometry to quantitative analysis.

Stoichiometry

Stoichiometry is a segment of chemistry that deals with the quantitative relationships of the elements and compounds as they undergo chemical transformations. In the case of the textbook problem, stoichiometry is implied through the use of molar masses to determine how much of a substance is reacting.

However, stoichiometry extends far beyond that, encompassing reactions where substances react in defined ratios. It can be used to predict yields, understand limiting reactants, and much more. The importance of stoichiometry lies in its ability to enable predictions about the outcomes of chemical reactions, which is fundamental in both academic and practical chemistry.

Dilution and Solution Preparation

Finally, the problem hints at the concept of dilution and solution preparation when it instructs to dissolve and then 'make up to' a certain volume. Dilution involves adding more solvent to a solution to decrease the concentration of the solute. This technique is routinely used in laboratories to prepare solutions of a desired molarity.

Understanding dilution is important, especially when working with concentrated solutions that may need to be diluted to a specific concentration for reactions or assays. The preparation of solutions with accurate concentrations is vital across various applications, from research experiments to the manufacturing of pharmaceuticals.