Question

An element of atomic mass 40 occur in fcc unit cell with edge length of 540 pm. If its density is 1.7 g cm-3 than calculate Avogadro number.​

Answer 1

D=M×z/Na×a³

M= atomic mass

Z=4 (FCC)

a=540 pm

D=1.7

Substitute and u'll get the answer

The answer is6.023×10^23

I hope this helps ( ╹▽╹ )

Answer 2

$\underline{\red{\mathrm{Answer:}}}$////$\bigstar$ Given://⇒ Edge length of the unit cell (a) = 540 pm = 540//⇒ No. of atoms per unit cell (Z) = 4//⇒ Density of the unit cell ($\rho$) = $1.7\; gm\;/ cm^3$//⇒ Atomic mass of the element (M) = $40\;g\;/mol$$\bigstar$ ////To Find:⇒ Avogadro number $N_0$$\bigstar$ ////Solution:We know that, //Density ($\rho$) is given by the formula,$\boxed{\mathsf{\rho = \dfrac{Z*M}{a^3*N_0*10^\ -30} }}$Therefo