Question

One \(g \mathrm{~m}\) metal \(\mathrm{M}^{3+}\) was discharged by the passage of \(1.81 \times 10^{23}\) electrons. What is the atomic weight of metal? (a) \(33.35\) (b) \(133.4\) (c) \(66.7\) (d) None of these

Short answer

The atomic weight of metal is (b) 133.4.

Step-by-step solution

- Understand the quantities given

One gram of a trivalent metal ion \(\mathrm{M}^{3+}\) is reduced by the passage of \(1.81 \times 10^{23}\) electrons. We need to find the atomic weight of the metal. The trivalency indicates that each ion accepts three electrons to become neutral.

- Calculate moles of electrons

First, calculate the moles of electrons using Avogadro's number \(6.022 \times 10^{23}\) electrons per mole. \[ \text{Moles of electrons} = \frac{1.81 \times 10^{23} \text{ electrons}}{6.022 \times 10^{23} \text{ electrons/mole}} \]

- Relate moles of electrons to moles of metal atoms

Since the metal ion is trivalent, each metal atom requires three electrons to become neutral. Therefore, \[ \text{Moles of metal atoms} = \frac{1}{3} \times \text{Moles of electrons} \]

- Calculate molar mass

Given that 1 gram of the metal was discharged, this corresponds to the moles of metal calculated in Step 3. The atomic weight (molar mass) of the metal can now be found by \[ \text{Atomic weight (Molar mass)} = \frac{\text{Mass of the metal in grams}}{\text{Moles of metal atoms}} \]

- Solve for the atomic weight

Perform the calculation by substituting the values obtained in previous steps to find the atomic weight of the metal. \[ \text{Atomic weight (Molar mass)} = \frac{1 \text{ g}}{\text{Moles of metal atoms from Step 3}} \]

Key concepts

Trivalent Metal Ion Reduction

Reduction is a chemical reaction that involves the gain of electrons by an element or compound. In the context of a trivalent metal ion like \(\mathrm{M}^{3+}\), reduction specifically refers to the process of gaining three electrons to achieve a neutral state. The 'tri' in trivalent denotes that each metal ion will require three electrons to complete this reduction process.

Understanding this concept is crucial when calculating the atomic weight of such a metal. The question from the textbook helps illustrate this: one gram of trivalent metal ion is reduced by \(1.81 \times 10^{23}\) electrons, which implies that a definite number of metal ions have accepted electrons, with each ion accepting three.

In practical terms, trivalent metal ions are often involved in electrochemical reactions such as battery operation or corrosion processes, where their ability to accept electrons plays a significant role in the chemistry of the system.

Mole Concept in Chemistry

The mole concept is a fundamental cornerstone in the field of chemistry, providing a bridge between the macroscopic world we observe and the microscopic world of atoms and molecules. One mole is defined as the amount of a substance that contains as many entities (atoms, molecules, ions, etc.) as there are atoms in 12 grams of carbon-12. The number of entities per mole is a fixed number, known as Avogadro's number, which is approximately \(6.022 \times 10^{23}\).

The mole is vital in translating quantities in chemistry. To find the atomic weight of a metal from the question given, we use the mole concept to relate the number of electrons transferred to the amount of metal reduced. Understanding how to convert electrons to moles allows chemists to calculate the quantities of substances involved in reactions quantitatively.

Avogadro's Number Application

Avogadro's number \(6.022 \times 10^{23}\) represents a cornerstone of quantitative chemistry. It serves as the scaling factor that translates a mole, an abstract yet essential chemical unit, into tangible amounts we can work with in the laboratory.

When we apply Avogadro's number to the exercise in question, we are essentially translating the number of electrons that have been used to reduce the metal ion into moles. Since each mole of electrons contains \(6.022 \times 10^{23}\) electrons, dividing the total number of electrons by Avogadro's number yields the moles of electrons. This step is critical to solving the problem as it links the given quantitative electron count to the theoretical framework of moles, enabling further calculations towards finding the atomic weight of the metal.

Stoichiometry

Stoichiometry is the section of chemistry that pertains to the calculation of the quantities of reactants and products in chemical reactions. It is founded on the conservation of mass and the concept of moles. In stoichiometry, the relationship between the quantities of reactants and products can be expressed through balanced chemical equations.

In the context of our exercise, stoichiometry guides the relationship between moles of electrons and moles of metal atoms. Since each metal atom requires three electrons to be reduced, a triad ratio is established: for every three moles of electrons, there is one mole of metal atoms. The ability to correctly equate the quantities involved in a chemical reaction, as shown in the step-by-step solution, leads to the determination of the metal's atomic weight, proving stoichiometry's pivotal role in solving chemical problems.