Problem 62 - Chemistry The Central Science

Question

(a) Calculate the molarity of a solution made by dissolving 12.5 grams of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) in enough water to form exactly \(550 \mathrm{~mL}\) of solution. (b) How many moles of KBr are present in \(150 \mathrm{~mL}\) of a \(0.275 \mathrm{M}\) solution? (c) How many milliliters of 6. \(1 \mathrm{M}\) HCl solution are needed to obtain \(0.100 \mathrm{~mol}\) of \(\mathrm{HCl} ?\)

Step-by-Step Solution

Answer

(a) The molarity of the \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) solution is approximately 0.1403 M. (b) There are 0.04125 mol of KBr in 150 mL of a 0.275 M solution. (c) Approximately 16.39 mL of 6.1 M HCl solution is needed to obtain 0.100 mol of HCl.

Step 1: a. Calculating the molarity of a Na2CrO4 solution

To calculate the molarity, we need to follow these steps: 1. Find the molar mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\). 2. Calculate the number of moles of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) dissolved. 3. Convert the volume of the solution from mL to L. 4. Divide the number of moles by the volume in liters to find the molarity. #Step 1: Finding the molar mass of Na2CrO4#

Step 1: Finding the molar mass of Na2CrO4

Using the periodic table, we can find the molar masses of each element and add them together: Molar mass of Na = 22.99 g/mol Molar mass of Cr = 51.996 g/mol Molar mass of O = 16.00 g/mol Molar mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) = (2 × 22.99) + 51.996 + (4 × 16.00) = 161.97 g/mol #Step 2: Calculating the number of moles of Na2CrO4 dissolved#

Step 2: Calculating the number of moles of Na2CrO4 dissolved

To find the number of moles, we divide the mass of the solute by its molar mass: Moles of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) = 12.5 g / 161.97 g/mol ≈ 0.07717 mol #Step 3: Converting the volume of the solution to liters#

Step 3: Converting the volume of the solution to liters

To convert the volume from mL to L, we can use the following relation: 1 L = 1000 mL 550 mL = 550 / 1000 = 0.550 L #Step 4: Calculating the molarity of the Na2CrO4 solution#

Step 4: Calculating the molarity of the Na2CrO4 solution

Finally, to find the molarity, we can use the formula: Molarity = Moles of solute / Volume of solution in liters Molarity of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) solution = 0.07717 mol / 0.550 L ≈ 0.1403 M #b. Finding the number of moles of KBr in a specific volume of a solution#

Step 6: b. Finding the number of moles of KBr in a specific volume of a solution

Given the volume and the molarity of the KBr solution, we can find the number of moles by multiplying them together. Moles of KBr = Molarity × Volume First, convert the volume given in mL to L: 150 mL = 150 / 1000 = 0.150 L Now, calculate the number of moles: Moles of KBr = 0.275 M × 0.150 L = 0.04125 mol #c. Finding the required volume of a solution to obtain a desired amount of solute#

Step 7: c. Finding the required volume of a solution to obtain a desired amount of solute

Given the molarity and the desired number of moles for the HCl solution, we can find the required volume by dividing the number of moles by the molarity. Volume = Moles / Molarity First, calculate the volume in liters: Volume of HCl solution = 0.100 mol / 6.1 M ≈ 0.01639 L Now, convert the volume to mL: 0.01639 L × 1000 = 16.39 mL So, approximately 16.39 mL of 6.1 M HCl solution is needed to obtain 0.100 mol of HCl.

Key Concepts

Molar Mass

Molar mass is a fundamental concept in chemistry, referring to the mass of one mole of a substance. It is measured in grams per mole (g/mol) and is the sum of the atomic masses of all the atoms in the molecule times the number of each type of atom. Understanding the molar mass is crucial for converting between mass and moles of a substance, a critical step in molarity calculations.

When calculating molar mass, students should remember to use the accurate atomic masses from the periodic table and multiply by the number of each element in the compound. For instance, in the exercise, the molar mass of \(\mathrm{Na}_2 \mathrm{CrO}_4\) was found by adding the atomic masses of sodium (Na), chromium (Cr), and oxygen (O) accordingly.

Moles of Solute

Moles of solute are a measure of the amount of a substance present in a solution. One mole corresponds to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities, be it atoms, molecules, or ions. To establish the moles of solute in a solution, divide the mass of the dissolved compound (in grams) by its molar mass.

For example, in the provided exercise, to find the number of moles of \(\mathrm{Na}_2 \mathrm{CrO}_4\), the mass of the compound (\(12.5 \text{ grams}\)) is divided by its computed molar mass (\(161.97 \text{ g/mol}\)). This calculation is the cornerstone of stoichiometry and molarity calculations, essential for understanding the composition of a solution.

Solution Volume

Solution volume is simply the amount of space that a solution occupies and is usually measured in liters (L) or milliliters (mL). In chemistry, most calculations require the volume to be in liters, so a conversion from milliliters to liters is often necessary, using the conversion factor that \(1 \text{ L} = 1000 \text{ mL}\).

In our exercise example, the solution volume for the \(\mathrm{Na}_2 \mathrm{CrO}_4\) solution is given in milliliters and is converted to liters by dividing by 1000. Students must ensure proper conversion to avoid errors in further calculations, such as determining molarity.

Stoichiometry

Stoichiometry involves the quantitative relationships of substances as they participate in chemical reactions. It allows chemists to predict the amounts of products and reactants involved in a reaction based on the balanced chemical equation. This concept extends to solutions where stoichiometry is used to determine molarity, moles of solute, and volume based on known quantities.

For instance, knowing the volume and molarity of a solution allows us to calculate the moles of solute, as seen in the KBr example in our exercise. Similarly, if we have moles and molarity, we can determine the solution volume required. Mastering stoichiometry is key to accurately conducting chemical calculations.