At 100, ^circ C,copper (Cu) has FCC unit cell | vedclass

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

Question

(D) The density $(d)$ of a unit cell is given by the formula: $d = \frac{Z 	imes M}{N_A 	imes a^3}$For an $FCC$ unit cell,the number of atoms per un

Vedclass · Accepted Answer

$\frac{422}{x^3}$

Vedclass · Answer

(D) The density $(d)$ of a unit cell is given by the formula: $d = \frac{Z 	imes M}{N_A 	imes a^3}$For an $FCC$ unit cell,the number of atoms per unit cell $(Z) = 4$.The atomic mass of $Cu$ $(M) = 63.55\, g\, mol^{-1}$.The Avogadro constant $(N_A) = 6.023 	imes 10^{23}\, mol^{-1}$.The edge length $(a) = x\, \mathring{A} = x 	imes 10^{-8}\, cm$.Substituting these values into the formula:$d = \frac{4 	imes 63.55}{(6.023 	imes 10^{23}) 	imes (x 	imes 10^{-8})^3}$$d = \frac{254.2}{6.023 	imes 10^{23} 	imes x^3 	imes 10^{-24}}$$d = \frac{254.2}{0.6023 	imes x^3} \approx \frac{422}{x^3}\, g\, cm^{-3}$

At $100\, ^\circ C$,copper $(Cu)$ has $FCC$ unit cell structure with cell edge length $x\, \mathring{A}$. What is the approximate density of $Cu$ (in $g\, cm^{-3}$) at this temperature? [Atomic mass of $Cu = 63.55\, u$]

Similar Questions

Calculate the edge length of a unit cell if a metal having an atomic radius of $170 \ pm$ forms a simple cubic unit cell.

An element (atomic mass $= 60$) has a face-centered cubic $(FCC)$ structure with a density of $6.23 \ g \ cm^{-3}$. The edge length of the unit cell is.... $(N_A = 6.02 \times 10^{23} \ mol^{-1})$

Calculate the volume of the unit cell for an element having a molar mass of $56 \ g \ mol^{-1}$ that forms $bcc$ unit cells. $\left[\rho \cdot N_{A} = 4.8 \times 10^{24} \ g \ cm^{-3} \ mol^{-1}\right]$

An element has $2.03 \times 10^{24}$ atoms in $135 \ g$. If the element crystallizes in a face-centered cubic $(FCC)$ lattice structure with an edge length of $150 \ pm$, then the density of the element is ............... $g \ cm^{-3}$.

Calculate the molar mass of a metal having a density of $9.3 \ g \ cm^{-3}$ that forms a simple cubic unit cell. $[a^3 \cdot N_A = 22.6 \ cm^3 \ mol^{-1}]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module