A diatomic molecule X_2 has a body-centred cu | vedclass

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

Question

(C) The density formula is $ho = \frac{Z 	imes M}{N_A 	imes a^3}$, where $Z = 2$ for a $bcc$ structure.Given $ho = 6.17 \ g \ cm^{-3}$, $a = 300

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) The density formula is $ho = \frac{Z 	imes M}{N_A 	imes a^3}$, where $Z = 2$ for a $bcc$ structure.Given $ho = 6.17 \ g \ cm^{-3}$, $a = 300 \ pm = 300 	imes 10^{-10} \ cm$, and $N_A = 6 	imes 10^{23} \ mol^{-1}$.Substituting the values: $6.17 = \frac{2 	imes M}{6 	imes 10^{23} 	imes (300 	imes 10^{-10})^3}$.$6.17 = \frac{2 	imes M}{6 	imes 10^{23} 	imes 27 	imes 10^{-24}} = \frac{2 	imes M}{16.2}$.$M = \frac{6.17 	imes 16.2}{2} \approx 50 \ g \ mol^{-1}$.Number of molecules $= \frac{	ext{mass}}{M} 	imes N_A = \frac{200 \ g}{50 \ g \ mol^{-1}} 	imes N_A = 4 \ N_A$.

$A$ diatomic molecule $X_2$ has a body-centred cubic (bcc) structure with a cell edge of $300 \ pm$. The density of the molecule is $6.17 \ g \ cm^{-3}$. The number of molecules present in $200 \ g$ of $X_2$ is (Avogadro constant $N_A = 6 \times 10^{23} \ mol^{-1}$) (in $N_A$)

Similar Questions

The crystal structure of an element has an $fcc$ lattice. If the edge length of the crystal is $4 \ \mathring{A}$,what is the atomic weight (in $g \ mol^{-1}$) of the element,if the density of the crystal is $11.21 \ g \ cm^{-3}$ $(N_{A} = 6.023 \times 10^{23} \ mol^{-1})$?

$Xe$ crystallizes in an $FCC$ structure. The edge length of its unit cell is $620 \, pm$. The radius of $Xe$ is $=$ ...... $pm$.

The density of the unit cell of $Li$ is $0.539 \ g/cm^3$ and the edge length of the unit cell is $3.5 \ \mathring{A}$. How many $Li$ atoms are present in a unit cell of lithium? (Atomic mass of $Li = 6.94 \ g/mol$)

$A$ compound forms a $bcc$ unit cell with an edge length of $400 \text{ pm}$. Determine the molar mass of the compound if the density of the compound is $3.5 \text{ g cm}^{-3}$. (Given: $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$)

In the Bragg's equation for diffraction of $X$-rays, $n$ represents:

Vedclass Test Series

Exam Paper Generator

Online Exam Module