Question

Gold crystalline in fcc lattice calculate no. Of unit cells in 2 mg of gold

Answer 1

Answer:

Explanation:

Molar mass of gold =197

No. Of atoms in 2g of gold

2/197*6.022*10^23

=6.1137*10^21

As the FCC contains 4 atoms

Unit cell=6.1137/4=1.528*10^21

Hence it contains. 1.528*10^21 cells

Answer 2

$\huge\underline\blue{\sf Answer:}$

$\red{\boxed{\sf No.\:of\:Unit\:Cell=1.55×10^{21}}}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Mass of gold = 2mg or $\sf{2×10^{-3}g}$

• In FCC lattice no. of Atoms = 4

$\large\underline\pink{\sf To\:Find: }$

• Number of unit cell = ?

━━━━━━━━━━━━━━━━━━━━━━━━━

Atomic mass of gold = 197

i.e 1 mole of gold $\implies{\sf 197 \implies 6.02×10^{23}}$

Hence,

No of Atoms present in 1g of gold =$\sf{\frac{2×10^{-3}×6.02×10^{23}}{197}}$

$\large\implies{\sf \frac{12.04×10^{23}}{197} }$

Now ,

No. of unit cell present in 2mg of Gold :-

$\large\implies{\sf \frac{12.04×10^{23}}{197×4} }$

$\large\implies{\sf \frac{12.04×10^{23}}{776} }$

$\large\implies{\sf 1.55×10^{21}}$

$\red{\boxed{\sf No.\:of\:Unit\:Cell=1.55×10^{21}}}$