Problem 35 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

If an iodized salt contains \(1 \%\) of \(\mathrm{KI}\) and a person takes \(2 \mathrm{~g}\) of the salt every day, the iodine ions going into his body everyday would be approximately \((\mathrm{K}=39\), \(\mathrm{I}=127\) ) (a) \(7.2 \times 10^{21}\) (b) \(7.2 \times 10^{18}\) (c) \(3.6 \times 10^{21}\) (d) \(9.5 \times 10^{18}\)

Step-by-Step Solution

Answer

None of the options match the correct number of iodine ions, which is approximately \(7.2 \times 10^{19}\), indicating an error in the provided options or in the calculation steps.

Step 1: Calculate the mass of KI in the daily intake

First, calculate the amount of potassium iodide (\text{KI}) in grams that the person takes every day. This can be done by finding 1% of the daily intake of salt which is 2 grams. The calculation is as follows: \(0.01 \times 2 \text{ g} = 0.02 \text{ g} \) of \( \text{KI} \).

Step 2: Calculate the number of moles of KI

Calculate the number of moles of potassium iodide (\text{KI}) using its molar mass. The molar mass of \text{KI} is the sum of the atomic masses of potassium (K) and iodine (I), which are 39 and 127 respectively. Thus, the molar mass of \text{KI} is \(39 + 127 = 166 \text{ g/mol}\). The number of moles of \text{KI} is then calculated by dividing the mass of \text{KI} by its molar mass: \(\frac{0.02 \text{ g}}{166 \text{ g/mol}} = 1.2048192771 \times 10^{-4} \text{ moles of KI}\).

Step 3: Calculate the number of iodine ions

Each mole of \text{KI} contains Avogadro's number of iodine ions, which is approximately \(6.022 \times 10^{23}\). To find the total number of iodine ions, multiply the number of moles of \text{KI} by Avogadro's number: \(1.2048192771 \times 10^{-4} \text{ moles} \times 6.022 \times 10^{23} \text{ ions/mole} = 7.25 \times 10^{19} \text{ iodine ions} \).

Step 4: Determine the correctness of given options

Considering significant figures, the calculated number of iodine ions should be presented with two significant figures, since the initial percentage given was in two significant figures. That means the correct answer will be \(7.2 \times 10^{19}\) iodine ions. However, this does not match any of the given options, indicating a possible mistake in our calculation or in the provided options.

Key Concepts

Molar Mass Calculation

Understanding the molar mass of a compound is crucial in stoichiometry as it relates to the mass of one mole of that substance. The molar mass is calculated by summing the atomic masses of all the atoms in the compound. For example, in the case of potassium iodide (KI), the molar mass is found by adding the atomic mass of potassium (K), which is 39 grams per mole, and the atomic mass of iodine (I), which is 127 grams per mole. Hence, the molar mass of KI is calculated to be:
\[ 39 \text{g/mol} + 127 \text{g/mol} = 166 \text{g/mol}. \]
When we have a certain mass of the substance, dividing this mass by the molar mass gives us the number of moles. In the provided exercise, you would calculate the number of moles of KI in a 2-gram salt intake using this molar mass.

Avogadro's Number

Avogadro’s number, approximately \(6.022 \times 10^{23}\), is a fundamental constant in chemistry that represents the number of particles, such as atoms, ions, or molecules, in one mole of a substance. This colossal number is pivotal when transitioning from the microscopic world of atoms to the macroscopic world that we can measure in labs. As seen in the exercise, once you find the number of moles of KI, you multiply it by Avogadro's number to find the total count of iodine ions. This step bridges the gap between the weight of a compound (in grams) and the number of constituent particles it contains.

For instance, if the calculated number of moles of KI is \(1.2048192771 \times 10^{-4}\), you find the total number of iodine ions by multiplying it by Avogadro’s number: \[ 1.2048192771 \times 10^{-4} \text{ moles} \times 6.022 \times 10^{23} = 7.25 \times 10^{19} \text{ iodine ions}.\]

Percent Composition

Percent composition is used to describe the percentage by mass of each element within a compound. In the context of our exercise, iodized salt contains 1% potassium iodide by mass. This means that for every 100 grams of the salt, there is 1 gram of potassium iodide. Calculating the percent composition can be essential for determining how much of a certain element is present in a given amount of compound.

For example, if a person consumes 2 grams of this iodized salt, the mass of KI, which is 1% of the salt, can be calculated as: \[ 1\% \times 2 \text{g} = 0.02 \text{g}. \]
This concept helps us find out how much of a specific element or compound is present in a sample, which is the first step in calculating the amount that relates to the number of particles or moles.