Number of atoms in 560 g of Fe (atomic mass = 56 g/mol) is
(a) twice that of 70 g N
(b) half that of

Q: Why didn't we multiply by Avogadro's number ($N_A$) to find the exact number of atoms?

Because we only need to compare the quantities. Since $N_A$ is a constant, multiplying all mole values by $N_A$ won't change their relative ratios. Comparing moles is mathematically equivalent and much faster!

Q: Why did we use the atomic mass of Nitrogen as 14 and not 28?

The question asks for the number of *atoms*, not molecules. Even though Nitrogen gas exists as $N_2$, 70 g of Nitrogen will contain the same number of Nitrogen *atoms* regardless of whether it's in atomic or molecular form. Moles of atoms = Total Mass / Atomic Mass.

Q: What if the question asked for the number of molecules instead?

If it asked for molecules, we would divide the given mass by the molecular mass (e.g., 28 for $N_2$, 2 for $H_2$). But Iron is a metal and exists as individual atoms in its lattice, so we always use its atomic mass.

Q: Is the atomic mass of Hydrogen 1 or 1.008?

For standard JEE calculations unless specified otherwise, we approximate the atomic mass of Hydrogen to 1 g/mol to make calculations simpler and faster.

Q: Can I solve this using unitary method?

Yes! For example, 56 g of Fe contains $N_A$ atoms. So, 560 g contains $10 N_A$ atoms. Similarly, 14 g of N contains $N_A$ atoms, so 70 g contains $5 N_A$ atoms. $10 N_A$ is twice of $5 N_A$. It's the exact same logic, just phrased differently!

Question

Number of atoms in $560 	ext{ g}$ of Fe (atomic mass = $56 	ext{ g mol}^{-1}$) is

Options:

(A) twice that of $70 	ext{ g}$ N

(B) half that of $20 	ext{ g}$ H

(C) Both (a) and (b)

(D) None of the above

Sigma Percentile · Accepted Answer

### The Power of Proportions

When faced with a question asking us to compare the **number of atoms**, our first instinct might be to calculate the exact number using Avogadro's number, $N_A$. However, in competitive exams like JEE, time is of the essence.

Since $N_A$ is a universal constant, the number of atoms is always **directly proportional** to the number of moles of atoms. Therefore, comparing the moles is mathematically identical to comparing the actual number of atoms, but it saves us from tedious calculations!

### Analyzing the Iron Sample

Let's start by finding out how much Iron we actually have in terms of moles. The formula for the number of moles of atoms is:

$$n = \frac{	ext{Given Mass}}{	ext{Atomic Mass}}$$

We are given $560 	ext{ g}$ of Iron (Fe), and its atomic mass is $56 	ext{ g/mol}$. Plugging these values into our formula:

$$n_{	ext{Fe}} = \frac{560}{56} = 10 	ext{ moles}$$

So, we have exactly $10 	ext{ moles}$ of Iron atoms. This is our benchmark. Now, we need to evaluate the given options against this value.

### Evaluating Option (a): Nitrogen

Option (a) asks us to look at $70 	ext{ g}$ of Nitrogen (N). To find the moles of Nitrogen atoms, we divide its given mass by its atomic mass ($14 	ext{ g/mol}$):

$$n_{	ext{N}} = \frac{70}{14} = 5 	ext{ moles}$$

Now, let's compare this with our benchmark. We have $10 	ext{ moles}$ of Fe and $5 	ext{ moles}$ of N.

$$10 = 2 	imes 5$$

This clearly shows that the number of atoms in $560 	ext{ g}$ of Fe is **twice** that of $70 	ext{ g}$ of N. Therefore, option (a) is a perfectly correct statement.

### Evaluating Option (b): Hydrogen

Next, let's examine option (b), which involves $20 	ext{ g}$ of Hydrogen (H). The atomic mass of Hydrogen is $1 	ext{ g/mol}$.

$$n_{	ext{H}} = \frac{20}{1} = 20 	ext{ moles}$$

Comparing this to our Iron benchmark of $10 	ext{ moles}$:

$$10 = \frac{1}{2} 	imes 20$$

We can see that $10$ is exactly **half** of $20$. This means the number of atoms in the Iron sample is half that of the Hydrogen sample. Thus, option (b) is also a correct statement.

### The Final Verdict

Since both statements (a) and (b) have proven to be mathematically sound, the most accurate choice provided in the question is option (c), which encompasses both valid observations. This problem beautifully illustrates how mastering the mole concept can turn a seemingly complex comparison into a simple mental math exercise!

Animated Solution for Chemistry - Basic Concepts in Chemistry: Number of atoms in $560 g$ of Fe (atomic mass = $56 g mol^{- 1}$ ) is

Select Answer:

Visualized Solution

\text{Concept}

\text{Formula}

\text{Moles of Fe}

\text{Moles of Fe}

\text{Option (a): Moles of N}

\text{Option (a): Moles of N}

\text{Comparing Fe and N}

\text{Option (b): Moles of H}

\text{Option (b): Moles of H}

\text{Comparing Fe and H}

\text{Conclusion}

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

The Power of Proportions

Analyzing the Iron Sample

Evaluating Option (a)

Evaluating Option (b)

The Final Verdict

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Animated Solution for Chemistry - Basic Concepts in Chemistry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Number of atoms in 560 g of Fe (atomic mass = 56 g mol−1) is

Select Answer:

Visualized Solution

\text{Concept}

\text{Formula}

\text{Moles of Fe}

\text{Moles of Fe}

\text{Option (a): Moles of N}

\text{Option (a): Moles of N}

\text{Comparing Fe and N}

\text{Option (b): Moles of H}

\text{Option (b): Moles of H}

\text{Comparing Fe and H}

\text{Conclusion}

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

The Power of Proportions

Analyzing the Iron Sample

Evaluating Option (a)

Evaluating Option (b)

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