A crystalline solid AB adopts NaCl type structure with edge length of unit cell as 745 pm and formula mass of 74.5 a.m.u. The density of the crystalline compound is

To find the density of the crystalline compound AB that adopts the NaCl type structure, we can use the formula for density, which is defined as mass per unit volume. In this case, we will calculate the density using the edge length of the unit cell and the formula mass of the compound.

Understanding the NaCl Structure

The NaCl structure is a face-centered cubic (FCC) lattice where each unit cell contains four formula units of the compound. This is important because it allows us to determine the total mass of the unit cell based on the formula mass of the compound.

Step-by-Step Calculation

• Determine the volume of the unit cell: The volume \( V \) of a cubic unit cell can be calculated using the formula:

V = a^3

where \( a \) is the edge length of the unit cell. Given that the edge length \( a \) is 745 pm (or \( 745 \times 10^{-12} \) m), we can calculate the volume:

V = (745 \times 10^{-12} \, \text{m})^3 = 4.13 \times 10^{-28} \, \text{m}^3

• Calculate the mass of the unit cell: Since the formula mass of AB is 74.5 a.m.u., we need to convert this mass into kilograms. The conversion factor is:

1 \, \text{a.m.u.} = 1.66 \times 10^{-27} \, \text{kg}

Thus, the mass \( m \) of the unit cell containing four formula units is:

m = 4 \times 74.5 \, \text{a.m.u.} \times 1.66 \times 10^{-27} \, \text{kg/a.m.u.} = 4.95 \times 10^{-25} \, \text{kg}

• Calculate the density: Now that we have both the mass and the volume, we can find the density \( \rho \) using the formula:

\rho = \frac{m}{V}

Substituting the values we calculated:

\rho = \frac{4.95 \times 10^{-25} \, \text{kg}}{4.13 \times 10^{-28} \, \text{m}^3} \approx 120.0 \, \text{kg/m}^3

Final Result

The density of the crystalline compound AB, which adopts the NaCl type structure, is approximately 120.0 kg/m³. This value reflects the compactness of the structure and the mass of the unit cell, providing insight into the material's properties.