Calculate the number of atoms in 5 g of a me | vedclass

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

Question

(A) For a simple cubic unit cell, the number of atoms per unit cell $(z)$ $= 1$. Edge length $(a)$ $= 336 \ pm = 3.36 	imes 10^{-8} \ cm$. Density $(

Vedclass · Accepted Answer

$1.4 	imes 10^{22}$

Vedclass · Answer

(A) For a simple cubic unit cell, the number of atoms per unit cell $(z)$ $= 1$. Edge length $(a)$ $= 336 \ pm = 3.36 	imes 10^{-8} \ cm$. Density $(d)$ $= \frac{z 	imes M}{a^3 	imes N_A}$, where $M$ is the molar mass and $N_A$ is Avogadro's number $(6.022 	imes 10^{23} \ mol^{-1})$. $9.4 = \frac{1 	imes M}{(3.36 	imes 10^{-8})^3 	imes 6.022 	imes 10^{23}}$. $M = 9.4 	imes (3.793 	imes 10^{-23}) 	imes 6.022 	imes 10^{23} \approx 214.65 \ g \ mol^{-1}$. Number of atoms in $5 \ g = \frac{	ext{mass}}{	ext{molar mass}} 	imes N_A = \frac{5}{214.65} 	imes 6.022 	imes 10^{23} \approx 1.4 	imes 10^{22}$ atoms.

Calculate the number of atoms in $5 \ g$ of a metal that crystallises in a simple cubic unit cell structure with an edge length of $336 \ pm$. (Density of the metal $= 9.4 \ g \ cm^{-3}$)

Similar Questions

Calculate the number of atoms in $5.4 \ g$ of a metal forming an $fcc$ structure,given that the unit cell volume $\left(a^3\right)$ multiplied by density $\left(\varrho\right)$ is $7.2 \times 10^{-22} \ g$.

An element is found to crystallize with $BCC$ structure having density $8.55 \ g \ cm^{-3}$. What is the edge length of the unit cell? (Atomic mass of element $= 93$)

An element crystallizes in an $fcc$ structure with an edge length of $200 \, pm$. If $200 \, g$ of this element contains $24 \times 10^{23}$ atoms,calculate its density in $g \, cm^{-3}$.

What is the relationship between the edge length $(a)$ and the radius of the atom $(r)$ in a simple cubic unit cell?

The distance between $Na^{+}$ and $Cl^{-}$ ions in solid $NaCl$ of density $2.165 \ g \ cm^{-3}$ is $........ \times 10^{-10} \ m$. (Nearest Integer)
(Given : $N_{A} = 6.02 \times 10^{23} \ mol^{-1}$,Molar mass of $NaCl = 58.5 \ g \ mol^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module