Problem 8 - A Complete Resource Book in Chemistry for JEE Main

Question

Boron has two stable isotopes, \(\mathrm{B}^{10}(19 \%)\) and \(\mathrm{B}^{11}\) (18\%) Find the average atomic weight of boron. (a) \(11.2\) (b) \(10.8\) (c) \(12.6\) (d) \(9.6\)

Step-by-Step Solution

Answer

The average atomic weight of boron is 10.8, option (b).

Step 1: Understanding Atomic Weight Calculation

To find the average atomic weight of an element with isotopes, multiply the atomic mass of each isotope by its relative abundance (expressed as a decimal), then add these values together. This will give the weighted average.

Step 2: Convert Percentage to Decimal

Isotope {B}0 has an abundance of 19%, which is 0.19 as a decimal. {B}0 has an abundance of 18%, which is 0.18 as a decimal. However, it's important to note that typically, natural boron consists almost entirely of 0{B} and 1{B}, and the percentages should reflect 100% total abundance, so check if there's any missing information or error in the data.

Step 3: Note Correction

The problem data seems to be incomplete as it doesn't sum up to 100% for natural occurring Boron (typically with about 80.1% 1{B} and 19.9% {B}). To calculate properly, assume typical abundances: 80.1% for 1{B} and 19.9% for {B}.

Step 4: Calculate Contribution of Each Isotope

Using the corrected common natural abundances: Calculate 10.0 (mass) 19.9% = 1.99, and 11.0 (mass) 80.1% = 8.811. Add these: 1.99 + 8.811 = 10.801.

Step 5: Find the Average Atomic Weight

Sum up the results from Step 4, which is 10.801, and round to significant figures if necessary. This is the weighted average atomic mass of Boron.

Step 6: Select the Correct Option

From the choices provided (a) 11.2, (b) 10.8, (c) 12.6, (d) 9.6, the closest match to 10.801 is option (b) 10.8.

Key Concepts

Boron Isotopes

Boron is a fascinating element because it is primarily composed of two stable isotopes:

• Boron-10 (written as \(\mathrm{B}^{10}\)), which has an atomic mass of approximately 10.
• Boron-11 (denoted \(\mathrm{B}^{11}\)), known for its atomic mass close to 11.

The existence of these isotopes means that boron's atomic weight is not a whole number but rather a weighted average of these isotopes. Each isotope contributes differently to boron's overall atomic mass based on its relative abundance in nature. This can lead to some interesting applications and implications in both chemistry and physics, where precise measurements of atomic weights are crucial.

Weighted Average

The concept of a weighted average is central to calculating atomic weights when dealing with isotopes. In this context, a weighted average takes into account the mass and abundance of each isotope. It allows you to combine these different contributions into one meaningful figure. In the case of boron, the weighted average helps determine what we see as "boron's atomic weight" on the periodic table. To calculate this, you:

• Multiply the atomic mass of each isotope by its relative abundance (as a decimal).
• Add these products together to get the weighted average.

This process ensures that more abundant isotopes have a proportionally greater impact on the final atomic weight.

Isotope Abundance

Isotope abundance refers to how common each isotope of an element is in nature, usually given as a percentage. For boron, the abundances of its isotopes are typically about 19.9% for \(\mathrm{B}^{10}\) and 80.1% for\(\mathrm{B}^{11}\). These percentages tell us how much each isotope contributes to the average atomic weight. To convert these into values useful for calculations, you rewrite the percentages as decimals:

• 19.9% becomes 0.199
• 80.1% becomes 0.801

Such conversions are fundamental for computing the weighted average, as only decimals can accurately reflect the proportions when calculating weights.

Atomic Mass Calculation

Calculating the atomic mass of elements like boron involves carefully applying the relative abundance of isotopes. The revised problem showed, using typical abundance values:

• The contribution of \(\mathrm{B}^{10}\) can be calculated as 10.0 (atomic mass) x 0.199 (abundance) = 1.99.
• The contribution from\(\mathrm{B}^{11}\) as 11.0 (atomic mass) x 0.801 (abundance) = 8.811.
• Adding these contributions yields a total of 10.801.

This calculation provides the average atomic mass of boron, often rounded to more practical figures such as 10.8. Understanding this calculation reinforces how isotopic differences impact the overall atomic weight, affecting many chemical and physical properties of elements.