The number of atoms in 8 g of sodium is x * 10^23. The value of x is ......... .
(Nearest integer)
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Q: Why do we divide by 23?

23 is the molar mass of sodium. Dividing the given mass by the molar mass gives us the number of moles, which is the standard way to count particles in chemistry.

Q: Can I use 6.022 instead of 6.02 for Avogadro's number?

The question explicitly provides $N_A = 6.02 \times 10^{23}$. Always use the values given in the problem to avoid slight variations in the final answer, especially in integer type questions.

Q: What does 'nearest integer' mean here?

It means you should round your final decimal answer to the closest whole number. For example, 2.09 rounds down to 2, while 2.6 would round up to 3.

Q: Is the atomic mass unit (u) the same as grams per mole?

Numerically, yes! The atomic mass of a single atom in atomic mass units (u) has the exact same numerical value as the mass of one mole of those atoms in grams (g/mol).

Q: Why do we multiply moles by Avogadro's number?

Think of a mole like a 'dozen'. Just as 1 dozen means 12 items, 1 mole means $6.02 \times 10^{23}$ items. To find the total items, you multiply the number of 'dozens' (moles) by the amount in one dozen (Avogadro's number).

Question

The number of atoms in 8 g of sodium is $x 	imes 10^{23}$. The value of $x$ is ……… .
(Nearest integer)
[Given : $N_A = 6.02 	imes 10^{23} 	ext{ mol}^{-1}$
Atomic mass of Na = $23.0 	ext{ u}$]

Sigma Percentile · Accepted Answer

### Visualizing the Atomic Realm

Imagine holding a small, metallic block of pure sodium in your hand. It weighs exactly $8 	ext{ g}$. To the naked eye, it's just a solid chunk of metal, but if we could zoom in, we would see a bustling metropolis of billions upon billions of tiny sodium atoms. Our mission in this problem is to act as atomic accountants and find out exactly how many atoms are packed inside this block.

To help us on this journey, we are given two crucial pieces of information. First, the molar mass of sodium is $23.0 	ext{ g mol}^{-1}$. Second, we have the famous Avogadro's number, $N_A = 6.02 	imes 10^{23} 	ext{ mol}^{-1}$, which acts as our universal counting constant.

### The Bridge Between Worlds: The Mole Concept

We cannot simply put atoms on a scale and count them one by one. Instead, we use the concept of the **mole** as a bridge between the macroscopic world (grams) and the microscopic world (atoms).

The first step is to find out how many moles of sodium we have. The number of moles ($n$) is simply the given mass ($W$) divided by the molar mass ($M$).

$$n = \frac{W}{M}$$

Once we know the number of moles, we can find the total number of atoms ($N$) by multiplying the moles by Avogadro's number ($N_A$). This gives us our master equation:

$$N = \frac{W}{M} 	imes N_A$$

### Crunching the Numbers

Now, let's carefully substitute our known values into the master equation. We plug in $8$ for the mass, $23$ for the molar mass, and $6.02 	imes 10^{23}$ for Avogadro's number.

$$N = \frac{8}{23} 	imes 6.02 	imes 10^{23}$$

Let's do the math step-by-step. First, multiplying the numerator values, $8 	imes 6.02$ gives us $48.16$. Now, we divide that by $23$.

$$N = \frac{48.16}{23} 	imes 10^{23}$$

$$N \approx 2.093 	imes 10^{23}$$

### The Final Polish: Finding $x$

The question states that the number of atoms is in the format $x 	imes 10^{23}$. By comparing this given format with our calculated result, we can clearly see that $x = 2.093$.

However, there is a catch! The question specifically asks for the **nearest integer**. When we look at $2.093$, the decimal part is less than $0.5$, which means we round down.

Therefore, $x$ rounds off to **$2$**. And that is our final, elegant answer!

Animated Solution for Chemistry - Basic Concepts in Chemistry: The number of atoms in 8 g of sodium is $x \times 1 0^{23}$ . The value of $x$ is ……… . (Nearest integer) [Given : $N_{A} = 6.02 \times 1 0^{23} mol^{- 1}$ Atomic mass of Na = $23.0 u$ ]

Enter Numerical Value:

Visualized Solution

Analyzing the Given Data

Formula for Number of Atoms

Substituting the Values

Calculating the Value

Comparing with the Given Format

Conceptual Extensions

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

Visualizing the Atomic Realm

The Bridge Between Worlds

Crunching the Numbers

The Final Polish

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