Answer to Question #183476 in General Chemistry for klein lyse

Question

A hypothetical alloy has a face-centered cubic unit cell. Its density is 10,470 kg/m³_, and its atomic weight is 106.8 g/mole.

a) What is the atomic radius of the alloy in cm?

b) What is the volume of the single atom of the alloy in cm³?

Answer 1

Convert the density into $g/cm3$

D=$10470kg/m3$

D therefore= $10.47g/cm3$

$d=(z×M)/(a3×Na)$

For a FCC $Z=4$

Therefore,

10.47g/cm3= $(4×106.8g/mol)÷(a3×6.022×10^23) a3=6.776×10^-22.$

But volume$= l3$

L= $3√vol$

$L=3√6.776×10^-22 L=8.79×10^-8cm$

$=8.79×10^-8×(10^10/10^2cm)$

=8.79∆

But, $C^2=a^2+a^2 ,C^2=2a^2$

$C^2=2×(8.79)^2$

$C^=√154.53$

$C=12.43∆$

And,

$r=c/4$

r=12.43/4

Ans for part( a)

$r=3.12$ ans.

(B)

Vol of a single atom $V=4/3πr^3$

$V=4/3×22/7×3.12^3$

$V=127.27cm^3$ Ans.