How many unit cells are present in a cube shaped ideal crystal of NaCl of mass 1.00 g? [Atomic mass

Q: Why is $Z=4$ for NaCl?

NaCl forms an FCC lattice where $Cl^-$ ions are at the corners and face centers, and $Na^+$ ions are at the edge centers and body center. Calculating the effective number of atoms yields 4 $Na^+$ and 4 $Cl^-$ ions, meaning there are 4 formula units of NaCl per unit cell.

Q: Why do we divide by Avogadro's number?

The molar mass ($58.5 \text{ g}$) is the mass of one *mole* ($6.02 \times 10^{23}$) of formula units. To find the mass of a *single* formula unit, we must divide the molar mass by Avogadro's number.

Q: Does the edge length '$a$' matter here?

No. While you could use the density formula $d = \frac{ZM}{N_A a^3}$ to find the volume of a unit cell and then divide the total volume by the unit cell volume, it's an unnecessarily long route. The mass method is direct and elegant, bypassing the need for volume or density entirely.

Q: What if the crystal was not ideal?

If the crystal had defects like Schottky defects (missing ions), the actual mass of a unit cell would be slightly less than the theoretical mass. Consequently, it would take slightly *more* unit cells to make up exactly 1 gram.

Q: Why did the question mention it was a 'cube shaped' crystal?

It's mostly descriptive filler and a slight distractor. The macroscopic shape of the crystal (whether it's a cube, a sphere, or an irregular shard) has absolutely no bearing on the number of microscopic unit cells contained within a specific mass.

Question

How many unit cells are present in a cube shaped ideal crystal of NaCl of mass $1.00 	ext{ g}$?
[Atomic mass : $	ext{Na} = 23$, $	ext{Cl} = 35.5$]

Options:

(A) $2.57 	imes 10^{21}$

(B) $5.14 	imes 10^{21}$

(C) $1.28 	imes 10^{21}$

(D) $1.71 	imes 10^{21}$

Sigma Percentile · Accepted Answer

### The Macroscopic vs. Microscopic World

Have you ever looked at a grain of table salt under a magnifying glass? It forms a perfect, beautiful little cube. But what happens if we zoom in a billion times more? We enter the microscopic world of the **unit cell**, the fundamental building block of the crystal.

In this problem, we are asked to bridge the gap between the macroscopic world (a $1.00 	ext{ g}$ crystal of NaCl) and the microscopic world (the number of unit cells). The question mentions a "cube-shaped" crystal, but don't let that distract you! The macroscopic shape is irrelevant; the only thing that dictates the number of unit cells is the **total mass**.

### Decoding the NaCl Unit Cell

Sodium chloride (NaCl) crystallizes in a **Face-Centered Cubic (FCC)** lattice. In this arrangement, the $Cl^-$ ions form the FCC lattice, and the $Na^+$ ions occupy all the octahedral voids.

What does this mean for our calculation? It means that a single, perfect unit cell of NaCl contains exactly **4 formula units** of NaCl. We denote this as $Z = 4$.

### The Mass of a Single Unit Cell

To find out how many unit cells are in $1.00 	ext{ g}$, we first need to know the mass of just *one* unit cell.

First, let's find the molar mass of NaCl:
$$M = 23 + 35.5 = 58.5 	ext{ g/mol}$$

This is the mass of one *mole* of NaCl formula units (which is Avogadro's number, $N_A = 6.02 	imes 10^{23}$, of units). But a single unit cell only contains $Z = 4$ formula units. Therefore, the mass of one unit cell ($m$) is the mass of these 4 units:
$$m = \frac{Z 	imes M}{N_A}$$

Substituting our values, we get the raw, uncalculated mass of a single unit cell:
$$m = \frac{4 	imes 58.5}{6.02 	imes 10^{23}} 	ext{ g}$$

### The Grand Calculation

Now, we have a total mass of $1.00 	ext{ g}$. If each unit cell weighs $m$ grams, the total number of unit cells ($N$) is simply the total mass divided by the mass of one unit cell:
$$N = \frac{	ext{Total Mass}}{m} = \frac{1}{m}$$

Let's plug in our expression for $m$. Notice how Avogadro's number beautifully flips to the numerator:
$$N = \frac{6.02 	imes 10^{23}}{4 	imes 58.5}$$

$$N = \frac{6.02 	imes 10^{23}}{234}$$

Dividing these numbers gives us our final answer:
$$N \approx 2.57 	imes 10^{21}$$

Take a moment to appreciate that number. In just one tiny gram of salt, there are over **two sextillion** unit cells perfectly stacked together. The sheer scale of Avogadro's number never fails to amaze!

Animated Solution for Chemistry - States of Matter: How many unit cells are present in a cube shaped ideal crystal of NaCl of mass $1.00 g$ ? [Atomic mass : $Na = 23$ , $Cl = 35.5$ ]

Select Answer:

Visualized Solution

\text{The NaCl Unit Cell}

\text{Molar Mass of NaCl}

\text{Mass of One Unit Cell}

\text{Substituting the Values}

\text{Number of Unit Cells}

\text{Final Calculation}

\text{The Result}

\text{Food for Thought}

The Sigma Insight: Solid State

The Macroscopic vs

Decoding the NaCl Unit Cell

The Mass of a Single Unit Cell

The Grand Calculation

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Animated Solution for Chemistry - States of Matter: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } How many unit cells are present in a cube shaped ideal crystal of NaCl of mass 1.00 g? [Atomic mass : Na=23, Cl=35.5]