The number of atoms present in 10.8 g of silver is: [atomic weight of silver =108 ]

Question

The number of atoms present in 10.8 g of silver is: [atomic weight of silver =108 ] (A)  6.123× 1023  (B)  6.312× 1022  (C)  6.023× 1022  (D)  6.076× 1023

Tardigrade Admin · Accepted Answer

∵ 108 g of silver contains =6.02 × 1023 atoms of Ag ∴ 10.8 g of silver contains =(6.023 × 1023/108) × 10.8 =6.023 × 1022 atoms

Q. The number of atoms present in $10.8 g$ of silver is :
[atomic weight of silver $= 108$ ]

$6.123 \times 10^{23}$

$6.312 \times 10^{22}$

$6.023 \times 10^{22}$

$6.076 \times 10^{23}$

$∵ 108 g$ of silver contains $= 6.02 \times 1 0^{23}$ atoms of $A g$
$∴ 10.8 g$ of silver contains $= \frac{6.023 \times 1 0 ^{23}}{108} \times 10.8$
$= 6.023 \times 1 0^{22}$ atoms

Q. The number of atoms present in 10.8g of silver is : [atomic weight of silver =108 ]

6.123×1023

6.312×1022

6.023×1022

6.076×1023