A substance forms a face-centered cubic (FCC) | vedclass

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

Question

(C) Given: $FCC$ structure, so $n = 4$. Density $(d)$ = $1.984 \ g \ cm^{-3}$. Edge length $(a)$ = $630 \ pm = 630 	imes 10^{-10} \ cm = 6.3 	imes 1

Vedclass · Accepted Answer

$74.70$

Vedclass · Answer

(C) Given: $FCC$ structure, so $n = 4$. Density $(d)$ = $1.984 \ g \ cm^{-3}$. Edge length $(a)$ = $630 \ pm = 630 	imes 10^{-10} \ cm = 6.3 	imes 10^{-8} \ cm$. Avogadro's number $(N_A)$ = $6.022 	imes 10^{23} \ mol^{-1}$. Using the formula: $d = \frac{n 	imes M}{a^3 	imes N_A}$. Rearranging for molar mass $(M)$: $M = \frac{d 	imes a^3 	imes N_A}{n}$. $M = \frac{1.984 	imes (6.3 	imes 10^{-8})^3 	imes 6.022 	imes 10^{23}}{4}$. $M = \frac{1.984 	imes 250.047 	imes 10^{-24} 	imes 6.022 	imes 10^{23}}{4}$. $M = \frac{298.81}{4} \approx 74.70 \ g \ mol^{-1}$.

$A$ substance forms a face-centered cubic $(FCC)$ crystal. Its density is $1.984 \ g \ cm^{-3}$ and the edge length of the unit cell is $630 \ pm$. Calculate the molar mass of the substance in $g \ mol^{-1}$.

Similar Questions

Lithium metal crystallises in a body-centred cubic crystal. If the length of the side of the unit cell of lithium is $351 \ pm,$ the atomic radius of lithium will be $............$ $pm$.

For a crystal, the angle of diffraction $(2 \theta)$ is $90^{\circ}$ and the second order line has a $d$ value of $2.28 \ \text{Å}$. The wavelength (in $\text{Å}$) of $X$-rays used for Bragg's diffraction is

Calculate the number of unit cells in $1 \text{ cm}^3$ of metal if it forms a simple cubic structure with a unit cell edge length of $500 \text{ pm}$.

How many unit cells are present in a cube-shaped ideal crystal of $NaCl$ of mass $1.00 \ g$?

Which of the following expressions is correct in the case of a $CsCl$ unit cell (edge length,$a$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module