An element with molar mass 2.7 times 10^{-2} | vedclass

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

Question

(D) The density formula is $d = \frac{z 	imes M}{N_A 	imes a^3}$.Given: $M = 2.7 	imes 10^{-2} \ kg \ mol^{-1}$,$a = 405 	imes 10^{-12} \ m$,$d =

Vedclass · Accepted Answer

$143$

Vedclass · Answer

(D) The density formula is $d = \frac{z 	imes M}{N_A 	imes a^3}$.Given: $M = 2.7 	imes 10^{-2} \ kg \ mol^{-1}$,$a = 405 	imes 10^{-12} \ m$,$d = 2.7 	imes 10^3 \ kg \ m^{-3}$,$N_A = 6.022 	imes 10^{23} \ mol^{-1}$.Substituting the values: $2.7 	imes 10^3 = \frac{z 	imes 2.7 	imes 10^{-2}}{6.022 	imes 10^{23} 	imes (405 	imes 10^{-12})^3}$.Solving for $z$: $z = \frac{2.7 	imes 10^3 	imes 6.022 	imes 10^{23} 	imes 6.643 	imes 10^{-29}}{2.7 	imes 10^{-2}} \approx 4$.Since $z = 4$,the unit cell is face-centered cubic $(fcc)$.For $fcc$,the relation between edge length $a$ and radius $r$ is $a = 2\sqrt{2}r$,so $r = \frac{a}{2\sqrt{2}}$.$r = \frac{405 	imes 10^{-12}}{2 	imes 1.414} = \frac{405 	imes 10^{-12}}{2.828} \approx 143.2 	imes 10^{-12} \ m$.Rounding to the nearest integer,$r = 143 	imes 10^{-12} \ m$.

An element with molar mass $2.7 \times 10^{-2} \ kg \ mol^{-1}$ forms a cubic unit cell with edge length $405 \ pm$. If its density is $2.7 \times 10^{3} \ kg \ m^{-3},$ the radius of the element is approximately......... $\times 10^{-12} \ m$ (to the nearest integer).

Similar Questions

Salt $AB$ has a zinc blende structure. The radii of $A^{2+}$ and $B^{2-}$ ions are $0.7 \ \mathring{A}$ and $1.8 \ \mathring{A}$ respectively. The edge length of the $AB$ unit cell is ........... $\mathring{A}$.

Calculate the edge length of a unit cell that crystallizes to form a $BCC$ structure. (Radius of atom is $2.17 \times 10^{-8} \ cm$,$\sqrt{3} = 1.732$)

If the edge length of a $bcc$ unit cell is $386 \ pm$, then the atomic radius will be ........... $pm$.

How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.

Calculate the number of atoms in $0.3 \ g$ of a metal if it forms a $bcc$ structure,given that $[\rho \times a^3 = 3 \times 10^{-22} \ g]$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module