Question

Iron oxide $FeO$, crystallises in a cubic lattice with a unit cell edge length of $50 \mathring{A}$ If density of the $FeO$ in the crystal is $40\, g \,cm ^{-3}$, then the number of $FeO$ units present per unit cell is ______ (Nearest integer) Given: Molar mass of $Fe$ and $O$ is $56$ and $16 \, g \, mol ^{-1}$ respectively $N _{ A }=60 \times 10^{23} mol ^{-1}$

Answer 1

Correct Answer: 4

To determine the number of $FeO$ units per unit cell, we start by using the formula for density (\( \rho \)) of a crystal lattice: \[ \rho = \frac{m \times n}{a^3 \times N_A} \] where \( m \) is the molar mass of $FeO$, \( n \) is the number of formula units per unit cell, \( a \) is the edge length of the unit cell, and \( N_A \) is Avogadro's number.

1. First, calculate the molar mass of $FeO$: \[ m = 56 + 16 = 72 \, g/mol \]
2. Convert the edge length from angstroms to centimeters: \[ a = 50 \, \mathring{A} = 50 \times 10^{-8} \, cm \]
3. Insert the known values into the density formula and solve for \( n \): \[ 40 = \frac{72 \times n}{(50 \times 10^{-8})^3 \times 6.02 \times 10^{23}} \]
4. Simplify and solve for \( n \): \[ 40 = \frac{72 \times n}{125 \times 10^{-24} \times 6.02 \times 10^{23}} \] \[ 40 = \frac{72 \times n}{7.525 \times 10^{-22}} \] \[ 72 \times n = 40 \times 7.525 \times 10^{-22} \] \[ n = \frac{301 \times 10^{-22}}{72} \] \[ n \approx 4.18 \]

Rounding to the nearest integer, the number of $FeO$ units per unit cell is 4. This value falls within the given range of 4-4.