The number of atoms of calcium in 40 text{ grams} of calcium is 6.02 * 10^(23). What is the mass,...

1. Translate the problem requirements: We know that 40 grams of calcium contains \(6.02 \times 10^{23}\) atoms. We need to find how many grams would contain \(1.806 \times 10^{24}\) atoms. This is asking us to scale up proportionally.
2. Establish the proportional relationship: Set up the relationship between mass and number of atoms using the given reference point as our baseline.
3. Calculate the scaling factor: Determine how many times larger the target number of atoms is compared to our reference amount.
4. Apply the scaling to find mass: Multiply the reference mass by the scaling factor to get our answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we know and what we're looking for in everyday terms.

We're told that 40 grams of calcium contains \(6.02 \times 10^{23}\) atoms. Think of this as our reference point - like knowing that one dozen eggs weighs a certain amount, and we want to figure out how much three dozen eggs would weigh.

We want to find: How many grams would contain \(1.806 \times 10^{24}\) atoms?

This is asking us to scale up proportionally. If we have more atoms, we'll have more mass in a predictable way.

Process Skill: TRANSLATE - Converting the scientific notation and chemistry context into a simple proportional relationship

2. Establish the proportional relationship

Let's think about this relationship in plain English first:

If 40 grams gives us \(6.02 \times 10^{23}\) atoms, then some unknown number of grams will give us \(1.806 \times 10^{24}\) atoms.

We can set this up as a proportion - like a recipe that we're scaling up or down:

40 grams is to \(6.02 \times 10^{23}\) atoms
as
X grams is to \(1.806 \times 10^{24}\) atoms

In mathematical notation (after our plain English understanding):
\(\frac{40 \text{ grams}}{6.02 \times 10^{23} \text{ atoms}} = \frac{X \text{ grams}}{1.806 \times 10^{24} \text{ atoms}}\)

3. Calculate the scaling factor

Now let's figure out how many times larger our target number of atoms is compared to our reference amount.

We need to compare \(1.806 \times 10^{24}\) to \(6.02 \times 10^{23}\).

Let's think about this step by step:

- \(1.806 \times 10^{24} = 18.06 \times 10^{23}\) (moving the decimal point)
- So we're comparing \(18.06 \times 10^{23}\) to \(6.02 \times 10^{23}\)

The scaling factor is: \(18.06 \div 6.02 = 3\)

This means we have exactly 3 times as many atoms as in our reference amount.

4. Apply the scaling to find mass

Since we have 3 times as many atoms, we'll have 3 times as much mass.

Our reference mass was 40 grams.

Therefore: \(3 \times 40 \text{ grams} = 120 \text{ grams}\)

Let's verify this makes sense: If 40 grams contains \(6.02 \times 10^{23}\) atoms, then 120 grams should contain \(3 \times 6.02 \times 10^{23} = 18.06 \times 10^{23} = 1.806 \times 10^{24}\) atoms ✓

Final Answer

The mass of \(1.806 \times 10^{24}\) atoms of calcium is 120 grams.

Looking at our answer choices, this matches choice D: 120.

Common Faltering Points

Errors while devising the approach

1. Confusing the given relationship
Students may misinterpret which quantity corresponds to which. They might think that \(6.02 \times 10^{23}\) grams contains 40 atoms, reversing the given relationship. This fundamental misreading leads to setting up the proportion incorrectly from the start.

2. Not recognizing this as a proportion problem
Some students may try to use complex chemistry formulas or atomic mass concepts instead of recognizing this as a straightforward proportional relationship. They overcomplicate what is essentially a scaling problem.

3. Setting up the proportion incorrectly
Even when students recognize it's a proportion, they may write it as: \(\frac{6.02 \times 10^{23}}{40} = \frac{1.806 \times 10^{24}}{X}\), putting atoms in the numerator instead of maintaining consistent units in numerator and denominator positions.

Errors while executing the approach

1. Scientific notation manipulation errors
When comparing \(1.806 \times 10^{24}\) to \(6.02 \times 10^{23}\), students often struggle with converting \(1.806 \times 10^{24}\) to \(18.06 \times 10^{23}\). They may incorrectly handle the powers of 10 or decimal point movement.

2. Division calculation mistakes
When calculating \(18.06 \div 6.02\), students may make arithmetic errors. Some might not recognize that \(18.06 = 3 \times 6.02\), leading to incorrect scaling factors like 2 or 4 instead of 3.

3. Cross-multiplication errors
If using cross-multiplication approach (\(\frac{40}{6.02 \times 10^{23}} = \frac{X}{1.806 \times 10^{24}}\)), students often make errors in handling the scientific notation during multiplication, especially when dealing with multiple powers of 10.

Errors while selecting the answer

1. Using the scaling factor as the final answer
After correctly calculating that the scaling factor is 3, some students might select an answer choice close to 3 or confuse the scaling factor with the final mass, forgetting to multiply by the original 40 grams.

2. Calculation verification errors
Students may arrive at 120 but second-guess themselves during verification. When checking if 120 grams gives \(1.806 \times 10^{24}\) atoms, they might make errors in the reverse calculation and switch to a different answer choice.