Question

Q.42 the number of atoms present in 10.8 g of silver is: (atomic
weight of silver = 108)
A) 6.123 x 1023
B) 6.312 x 1022
C) 6.023 x 1022
D) 6.076 x 1023​

Answer 1

Answer:

The answer is 6.023× 10^22

Explanation:

• One mole of atoms contain Avagadro number(6.023 ×10^23) of atoms.
• 10.8 g of silver is equal to 0.1 mole of atoms
• Therefore it has 6.023×10^22 number of silver atoms.
• The formula used is as follows:

$number \: of \: moles = \frac{weight}{atomic \:weight}$