Question

Total no. of electrons present in \(48 \mathrm{~g} \mathrm{Mg}^{2+}\) are: (a) \(24 N_{A}\) (b) \(2 N_{A}\) (c) \(20 \mathrm{~N}_{A}\) (d) none of these

Short answer

\(2 N_{A}\)

Step-by-step solution

Determine the molar mass of magnesium

Find the molar mass of magnesium (Mg) from the periodic table which is approximately 24 g/mol.

Calculate the number of moles of magnesium

Calculate moles of Mg using the formula: moles = given mass / molar mass. Given mass is 48 g of Mg.

Identify the charge of magnesium ions

Each magnesium ion (Mg²⁺) has lost two electrons, so each mole of Mg²⁺ ions would have lost two moles of electrons.

Calculate the total number of electrons lost

Since 1 mole of any substance contains Avogadro's number \(N_A\) of particles, 2 moles would contain \(2 \times N_A\) of electrons. Multiply the number of moles of Mg²⁺ found in Step 2 by 2 \(N_A\) to find the total number of electrons lost.

Key concepts

Avogadro's Number

Avogadro's number, denoted as \(N_A\), is a fundamental concept in chemistry that is pivotal when dealing with molecules and atoms at the microscopic scale. It refers to the number of units in one mole of any substance and is approximately equal to \(6.022 \times 10^{23}\). To illustrate, a mole of magnesium ions, which can be likened to a dozen of eggs referring to the count of 12, represents exactly \(N_A\) magnesium ions. In the context of chemical calculations, this constant allows us to transition from the microscopic world of individual atoms and electrons to the macroscopic world we can measure in a lab. For the given exercise, employing Avogadro's number is essential to find out the total number of electrons in \(48 \mathrm{g}\) of \(\mathrm{Mg}^{2+}\) ions.

Molar Mass

Molar mass is a property of substances that links the mass of a sample to the number of moles present. It is essentially the mass of one mole of a substance usually expressed in grams per mole (g/mol). The presence of a periodic table is essential as it provides the molar mass of individual elements like magnesium (\(\mathrm{Mg}\)). For the exercise, knowing the molecular mass of magnesium, which is about \(24 \mathrm{g/mol}\), allows us to calculate how many moles are present in a \(48 \mathrm{g}\) sample. To determine the moles of \(\mathrm{Mg}^{2+}\), you divide the given mass by the molar mass, thereby linking the sample's macroscopic mass to the microscopic quantity, moles, a pivotal step in stoichiometry.

Stoichiometry

Stoichiometry is like the recipe for a chemical reaction, showing the relationship between the reactants and the products. In terms of calculations, it lets us quantify the relationships in a balanced chemical equation, involving moles, mass, and molar ratios. In our electron count exercise, stoichiometry comes into play after determining the number of moles of \(\mathrm{Mg}^{2+}\). We must then consider the ionic charge, which indicates that each magnesium ion has lost two electrons. Stoichiometry allows us to deduce the total number of electrons lost (or gained) during ion formation, using the known ratios. It helps in establishing clear-cut proportions, such as one mole of \(\mathrm{Mg}^{2+}\) ions corresponding to two moles of electrons, which is crucial for accurate chemical calculations.

Chemical Calculations

Chemical calculations are the backbone of quantitative chemistry, providing the bridge between theoretical concepts and practical measurements. They often involve a series of conversions and use principles like the conservation of mass and charge. In our sample problem, several chemical calculations are needed, such as converting mass to moles, and using Avogadro's number to convert moles into particles, specifically electrons in this case. The final step multiplies the moles of \(\mathrm{Mg}^{2+}\) ions by \(2 \times N_A\), yielding the total number of lost electrons, as each ion loses two electrons. This kind of calculation is key to understanding topics ranging from reaction stoichiometry to electrochemistry, making it a powerful tool for students to master.