Question

an ax ceramic compound has the rock salt crystal structure. if the radii of the a and x ions are 0.137 and 0.241 nm, respectively, and the respective atomic weights are 22.7 and 91.4 g/mol, what is the density (in g/cm³) of this material?
0.438 g/cm³
1.75 g/cm³
0.175 g/cm³
4.38 g/cm³
question 2
2 pts

Explanation:

Step1: Determine the unit - cell type

For a rock - salt (face - centered cubic, FCC) structure of AX ceramic, the number of formula units per unit - cell, $n = 4$.

Step2: Calculate the edge - length of the unit - cell

The relationship between the radii of cations ($r_A$) and anions ($r_X$) in a rock - salt structure is $a = 2(r_A + r_X)$. Given $r_A=0.137\ nm$ and $r_X = 0.241\ nm$, then $a=2(0.137 + 0.241)\ nm=2\times0.378\ nm = 0.756\ nm=7.56\times10^{- 8}\ cm$.

Step3: Calculate the volume of the unit - cell

The volume of a cubic unit - cell, $V=a^{3}$. So $V=(7.56\times10^{-8}\ cm)^{3}=4.32\times10^{-22}\ cm^{3}$.

Step4: Calculate the molar mass of the compound

The molar mass of the compound $M = M_A+M_X$, where $M_A = 22.7\ g/mol$ and $M_X=91.4\ g/mol$. So $M=(22.7 + 91.4)\ g/mol=114.1\ g/mol$.

Step5: Calculate the density

The density formula is $
ho=\frac{nM}{V N_A}$, where $n = 4$, $M = 114.1\ g/mol$, $V=4.32\times10^{-22}\ cm^{3}$, and $N_A=6.022\times10^{23}\ mol^{-1}$.
$
ho=\frac{4\times114.1\ g/mol}{4.32\times10^{-22}\ cm^{3}\times6.022\times10^{23}\ mol^{-1}}$
$
ho=\frac{456.4\ g}{260.15\times10^{-1}\ cm^{3}}=\frac{456.4\ g}{26.015\ cm^{3}}\approx1.75\ g/cm^{3}$

Answer:

$1.75\ g/cm^{3}$