1.

Calculate number of atoms present in:(i) 0.1 mole of sulphur.(ii) 2 moles of phosphors.

Answer»

(i) 1 mole = 6.022 × 1023

0.1 mole = 0.1 × 6.022 × 1023 = 6.022 × 1022 molecules

1 mole of sulphur (S8) contains = 8 atoms

Hence, no of atoms present = 8 × no. of molecules

⇒ Number of atoms present = 8 × 6.022 × 1022 molecules

⇒ Number of atoms present = 4.8 × 1023 atoms

Thus, number of atoms present in 0.1 mole of sulphur is 4.8 × 1023 atoms.

(ii) 1 mole = 6.022 × 1023

2 mole = 2 × 6.022 × 1023 molecules

1 mole of phosphorus (P4) contains = 4 atoms

Hence, no of atoms present = 4 × no. of molecules

⇒ Number of atoms present = 4 × 2× 6.022 × 1023 molecules

⇒ Number of atoms present = 4.8 × 1024 atoms

Thus, number of atoms present in 2 mole of phosphorus is 4.8 × 1024 atoms.



Discussion

No Comment Found