Question

How many moles of \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) are present in \(225.0 \mathrm{~mL}\) of a \(1.20 \mathrm{M}\) solution? How many moles of \(\mathrm{Mg}^{2+}\) ions are present? How many moles of \(\mathrm{NO}_{3}{ }^{-}\)ions are present?

Short answer

0.27 mol of \(Mg(NO_3)_2\), 0.27 mol of \(Mg^{2+}\) ions and 0.54 mol of \(NO_3^{-}\) ions are present.

Step-by-step solution

Calculate moles of \(Mg(NO_3)_2\)

First let’s convert the volume of solution to liters as we know, molar concentration is defined in moles per liter: \(225.0 mL = 225.0/1000 = 0.225 L\). moles of \(Mg(NO_3)_2\) can be found using the formula \(Molarity = moles/volume\). Here, Molarity is given as \(1.20 M\) and volume is \(0.225 L\). Hence, the moles of \(Mg(NO_3)_2\) can be found by \(Moles = Molarity * Volume = 1.20 mol/L * 0.225 L = 0.27 mol\)

Determine moles of \(Mg^{2+}\) ions

One formula unit of \(Mg(NO_3)_2\) contributes one \(Mg^{2+}\) ion. So moles of \(Mg(NO_3)_2\) will be equal to moles of \(Mg^{2+}\) ions. The moles of \(Mg^{2+}\) ions present in the solution will be equal to \(0.27 mol\).

Calculate moles of \(NO_3^{-}\) ions

One formula unit of \(Mg(NO_3)_2\) contributes two \(NO_3^{-}\) ions. Hence, moles of \(NO_3^{-}\) will be twice the moles of \(Mg(NO_3)_2\). So, \(Moles = 2*0.27 mol = 0.54 mol\).

Key concepts

Mole Concept and Its Importance

Understanding the mole concept is vital in chemistry. A mole is a unit used to measure the amount of a substance, similar to how a dozen refers to twelve items. Specifically, one mole equates to Avogadro's number, which is approximately 6.022 x 10^23 particles - these could be atoms, molecules, or ions.

For instance, in the given problem, we're asked about the number of moles of \(Mg(NO_3)_2\), \(Mg^{2+}\) ions, and \(NO_3^{-}\) ions. This understanding allows chemists to count particles in a given sample by weighing it, which is practical given that counting individual atoms or molecules is impossible on a macroscopic scale. It's essential to grasp this concept as it is the foundation of stoichiometry and concentration calculations in solutions.

Calculating the Concentration of a Solution

The concentration of a solution is a measure of how much solute is dissolved in a given volume of solvent. Molarity, denoted as \(M\), is one of the most common units of concentration and is defined as moles of solute per liter of solution.

Let's break down the problem: we are given a solution with a volume of 225.0 mL and a concentration of 1.20 M. To find the moles of \(Mg(NO_3)_2\) present, convert the volume from milliliters to liters since molarity is per liter. Following that, you multiply the volume in liters by the concentration. Put simply, the formula \(Moles = Molarity \times Volume\) is what enables us to find the answer. It's an easy yet powerful equation for determining the composition of solutions.

Stoichiometry and its Applications

Stoichiometry is a section of chemistry that involves calculating the quantities of reactants and products involved in a chemical reaction. It's based on the conservation of mass and the concept of moles. In the context of our problem, stoichiometry helps determine the relationship between the compounds and their constituent ions.

Each \(Mg(NO_3)_2\) formula unit, for example, yields one \(Mg^{2+}\) ion and two \(NO_3^{-}\) ions. Because one mole of \(Mg(NO_3)_2\) contains Avogadro's number of formula units, you'd have an equal amount (in moles) of \(Mg^{2+}\) ions and twice that amount for \(NO_3^{-}\) ions. This ratio, derived from the chemical formula, guides you to accurately calculate the amount of individual ions present in a solution.