In a cubic unit cell of an ionic compound, th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) $1$. Identify the number of formula units per unit cell $(Z)$: The anions are at the $8$ corners, contributing $8 	imes \frac{1}{8} = 1$ anion. T

Vedclass · Accepted Answer

$7.0 	imes 10^{-23} 	ext{ cm}^3$

Vedclass · Answer

(C) $1$. Identify the number of formula units per unit cell $(Z)$: The anions are at the $8$ corners, contributing $8 	imes \frac{1}{8} = 1$ anion. The cation is at the body center, contributing $1$ cation. Thus, $Z = 1$.$2$. Use the density formula: $d = \frac{Z 	imes M}{N_A 	imes V}$, where $V$ is the volume of the unit cell.$3$. Rearrange to solve for $V$: $V = \frac{Z 	imes M}{d 	imes N_A}$.$4$. Substitute the values: $V = \frac{1 	imes 168.6 	ext{ g mol}^{-1}}{4 	ext{ g cm}^{-3} 	imes 6.022 	imes 10^{23} 	ext{ mol}^{-1}}$.$5$. Calculate: $V = \frac{168.6}{24.088 	imes 10^{23}} \approx 7.0 	imes 10^{-23} 	ext{ cm}^3$.

Similar Questions

Tobacco seed globin forms face-centered cubic crystals with a unit cell dimension of $12.3 \, nm$ and a density of $1.287 \, g \, cm^{-3}$. Its molecular mass will be:

An element occurs in the body-centred cubic $(BCC)$ structure with an edge length of $288 \ pm$. The density of the element is $7.2 \ g \ cm^{-3}$. The number of atoms present in $208 \ g$ of the element is nearly:

Lithium crystallises into a body-centered cubic $(BCC)$ structure. What is the radius of lithium if the edge length of its unit cell is $351 \ pm$ (in $pm$)?

The number of atoms in $4.5 \ g$ of a face-centred cubic crystal with edge length $300 \ pm$ is (Given: Density $= 10 \ g \ cm^{-3}$ and $N_A = 6.022 \times 10^{23}$)

The distance between two parallel $(111)$ planes in a crystal with an edge length of $300 \, pm$ is ................ $pm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module