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

(A) Given:Density, $d = 2.7 	imes 10^{3} \,kg \,m^{-3}$Molar mass, $M = 2.7 	imes 10^{-2} \,kg \,mol^{-1}$Edge length, $a = 405 \,pm = 4.05 	imes 1

Vedclass · Accepted Answer

Face-centred cubic $(fcc)$

Vedclass · Answer

(A) Given:Density, $d = 2.7 	imes 10^{3} \,kg \,m^{-3}$Molar mass, $M = 2.7 	imes 10^{-2} \,kg \,mol^{-1}$Edge length, $a = 405 \,pm = 4.05 	imes 10^{-10} \,m$Avogadro's number, $N_A = 6.022 	imes 10^{23} \,mol^{-1}$Using the formula for density of a unit cell:$d = \frac{z \cdot M}{a^3 \cdot N_A}$Rearranging for $z$:$z = \frac{d \cdot a^3 \cdot N_A}{M}$Substituting the values:$z = \frac{(2.7 	imes 10^{3}) \cdot (4.05 	imes 10^{-10})^3 \cdot (6.022 	imes 10^{23})}{2.7 	imes 10^{-2}}$$z = \frac{2.7 	imes 10^{3} \cdot 66.43 	imes 10^{-30} \cdot 6.022 	imes 10^{23}}{2.7 	imes 10^{-2}}$$z \approx 4$Since the number of atoms per unit cell is $4$, the unit cell is face-centred cubic $(fcc)$.

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 $2.7 \times 10^{3} \,kg \,m^{-3}$, what is the nature of the cubic unit cell?

Similar Questions

The edge length of a body-centered cubic $(BCC)$ unit cell is $390 \ pm$. If the radius of the cation is $150 \ pm$, what is the radius of the anion? (in $pm$)

Ferrous oxide has a cubic structure and each edge of the unit cell is $5.0 \ \mathring{A}$. Assuming the density of the oxide is $4.0 \ g \ cm^{-3}$,calculate the number of $Fe^{2+}$ and $O^{2-}$ ions present in each unit cell ($M_w$ of $FeO = 72 \ g/mol$).

The density of a body-centered cubic $(BCC)$ crystal of Molybdenum is $10.3 \ g \ cm^{-3}$. Calculate the edge length of the unit cell in $pm$. (Atomic mass of $Mo = 95.94 \ g \ mol^{-1}$) (in $.9$)

Calculate the number of atoms in $100 \, g$ of a crystal that crystallizes in an $fcc$ structure, given that its density is $10 \, g/cm^3$ and the edge length is $200 \, pm$.

Calculate the number of atoms present in a unit cell for an element having a molar mass of $23 \ g \ mol^{-1}$ and a density of $0.96 \ g \ cm^{-3}$. Given that $a^3 \cdot N_{A} = 48 \ cm^3 \ mol^{-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module