Question

(a) Calculate the molarity of a solution made by dissolving \(0.750\) grams of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in enough water to form exactly \(850 \mathrm{~mL}\) of solution. (b) How many moles of \(\mathrm{KMnO}_{4}\) are present in \(250 \mathrm{~mL}\) of a \(0.0475 \mathrm{M}\) solution? (c) How many milliliters of \(11.6 \mathrm{M} \mathrm{HCl}\) solution are needed to obtain \(0.250 \mathrm{~mol}\) of \(\mathrm{HCl}\) ?

Short answer

(a) The molarity of the Na2SO4 solution is 0.00621 M. (b) There are 0.0119 moles of KMnO4 in the solution. (c) 21.6 mL of the 11.6 M HCl solution is needed to obtain 0.250 mol of HCl.

Step-by-step solution

(a) Calculating the moles of Na2SO4 and the molarity of the solution.

First, we need to find the moles of Na2SO4 using the formula: moles = mass / molecular weight. We are given the mass (0.750 g) of Na2SO4. The molecular weight of Na2SO4 can be found by summing the atomic weights of all the atoms in the formula: Molecular weight of Na2SO4 = (2 * Atomic weight of Na) + (1 * Atomic weight of S) + (4 * Atomic weight of O) = (2 * 22.99) + (32.07) + (4 * 16.00) = 45.98 + 32.07 + 64.00 = 142.05 g/mol. Now, we can find the moles of Na2SO4: moles = 0.750 g / 142.05 g/mol = 0.00528 mol. The volume of the solution is given as 850 mL, which is equal to 0.850 L. Now, we can find the molarity of the solution: M = moles / volume = 0.00528 mol / 0.850 L = 0.00621 M. So, the molarity of the Na2SO4 solution is 0.00621 M.

(b) Calculating the moles of KMnO4.

We are given the molarity of the KMnO4 solution (0.0475 M) and its volume (250 mL, or 0.25 L). We can find the moles of KMnO4 by rearranging the molarity formula: moles = M * volume = 0.0475 M * 0.25 L = 0.0119 mol. So, there are 0.0119 moles of KMnO4 in the solution.

(c) Finding the volume of HCl solution needed to obtain 0.250 mol of HCl.

We are given the molarity of the HCl solution (11.6 M) and need to obtain 0.250 mol of HCl. We can find the volume needed by rearranging the molarity formula: volume = moles / M = 0.250 mol / 11.6 M = 0.0216 L. To convert the volume into milliliters, multiply it by 1000: 0.0216 L * 1000 = 21.6 mL. So, 21.6 mL of the 11.6 M HCl solution is needed to obtain 0.250 mol of HCl.

Key concepts

Calculating Moles

Understanding how to calculate the number of moles in a substance is foundational in chemistry. Let's simplify the concept: a mole is a unit that measures the amount of substance. It's like counting apples in a basket, but instead, we count atoms or molecules in a chemical sample.

To calculate moles, we use the formula:
\[ \text{moles} = \frac{\text{mass}}{\text{molecular weight}} \]
The mass is how heavy your chemical sample is, and the molecular weight (or molar mass) is the sum of the atomic weights of all atoms in a molecule, measured in grams per mole (g/mol). Here's a practical example: if we have 0.750 grams of sodium sulfate (\(\mathrm{Na}_{2}\mathrm{SO}_{4}\)), and knowing that the molecular weight of sodium sulfate is 142.05 g/mol, we do a simple division: \(0.750 \text{g} \div 142.05 \text{g/mol}\), resulting in approximately 0.00528 moles of sodium sulfate.

In the problem's context, we use this understanding to determine how many moles are present in given samples of different substances, which is the crucial first step in many stoichiometric calculations.

Converting Grams to Moles

Mass-Mole Relationship

The bridge between the tangible weight of a sample and the abstract world of chemical formulas is the grams to moles conversion. To convert grams to moles, you simply divide the mass of the substance (in grams) by its molecular weight (in g/mol).

This relationship helps scientists and students to work with comprehensible quantities rather than dealing with billions and billions of individual atoms or molecules. For instance, if you have a 0.750 g sample of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\), with a molecular weight of 142.05 g/mol, the number of moles is found using the conversion:
\[ \text{moles} = \frac{0.750 \text{ g}}{142.05 \text{ g/mol}} = 0.00528 \text{ moles} \]
This straightforward calculation is essential in all quantitative aspects of chemistry, including stoichiometry, reaction yields, and solution preparation.

Dilution of Solutions

Dilution is like adding water to your orange juice to decrease its concentration. In chemistry, dilution involves adding more solvent to a solution to decrease the concentration of solutes. The core principle is that the amount of solute stays the same, but the total volume of the solution increases, leading to a lower concentration or molarity.

A handy equation to remember for dilution is:\[ M_1V_1 = M_2V_2 \]
In this equation, \(M_1\) and \(V_1\) are the initial molarity and volume, and \(M_2\) and \(V_2\) are the molarity and volume after dilution. When you perform dilution calculations, make sure you keep track of the units! Volumes should be in liters to be consistent with the molarity unit of moles/liter. By conserving the product of concentration and volume, we can determine how to prepare solutions of desired concentrations from more concentrated stock solutions, which is a routine task in laboratories.

Molarity Formula

The Heart of Solutions Chemistry

Molarity, often symbolized by the letter M, measures the concentration of a solution. It tells us how many moles of solute are present in a liter of solution. Let's decode the molarity formula:
\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
Simply put, if you dissolve a certain number of moles of a chemical into a certain volume of liquid, you get a solution with a certain molarity.

For instance, if 0.00528 moles of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) are dissolved to make 0.850 liters of solution, the molarity is: \[ M = \frac{0.00528 \text{ moles}}{0.850 \text{ liters}} = 0.00621 \text{M} \]
This concept is fundamental for mixing solutions with precise concentrations, an essential skill in both academic and professional chemistry settings. It enables scientists to replicate experimental conditions accurately or to scale reactions appropriately.