A metallic element exists in cubic lattice. Each edge of unit cell is 4 A o The density of metal is 6 .25 g /m 3 . How many unit cells will be present in 100 g of metal?

d = weight Total volume 6.25 = 100 n × 4 × 10 − 10 3 100 6.25 × 64 × 10 − 30 n = 2 .5 × 10 29

Chemistry - Solid State Question with Solution | TestHub

ChemistrySolid StateGeneralEasy2 minPYQ_2020

ChemistryEasysingle choice

A metallic element exists in cubic lattice. Each edge of unit cell is $4 \overset{o}{A}$ The density of metal is $6 .25 g {/m}^{3} .$ How many unit cells will be present in $100 g$ of metal?

Options:

Answer:

Solution:

$d = \frac{weight}{Total volume}$
$6.25 = \frac{100}{n \times {(4 \times 10^{- 10})}^{3}}$
$\frac{100}{6.25 \times 64 \times 10^{- 30}}$
$n = 2 .5 × 10^{29}$

Stream:NTA_ABHYASSubject:ChemistryTopic:Solid StateSubtopic:General

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...