Which has maximum number of atoms?
(1) 10.8 g of Ag(108)
(3) 5.6 g of Fe(56)
(2) 2.4 g of C(12)
(4) 54 g OF AL
i want with explanations
Answers
Answer:
4) 54g of Aluminum has highest number of atoms.
Here,
n = No. of moles
N = No.of Atoms
m = Mass
M = Molar Mass
N(0) = Avogadro No.
Important Formulas:
n = m/M
N = n x N(0)
Explanation:
1) 10.8g of Ag(108)
No. of atoms = No. of moles x Avogadro No.
=> N = n x N(0)
Since, n = m/M
=> N = (m/M) x N(0)
=> N = (10.8/108) x 6.022 x 10^23
=> N = 0.1 x 6.022 x 10^23
=> N = 0.6022 x 10^23
=> N = 6.022 x 10^22
2) 5.6g of Fe(56)
No. of atoms = No. of moles x Avogadro No.
=> N = n x N(0)
Since, n = m/M
=> N = (m/M) x N(0)
=> N = (5.6/56) x 6.022 x 10^23
=> N = 0.1 x 6.022 x 10^23
=> N = 6.022 x 10^22
3) 2.4g of C(12)
No. of atoms = No. of moles x Avogadro No.
=> N = n x N(0)
Since, n = m/M
=> N = (m/M) x N(0)
=> N = (2.4/12) x 6.022 x 10^23
=> N = 0.2 x 6.022 x 10^23
=> N = 1.2044 x 10^23
=> N = 12.044 x 10^22
4) 54g of Al(27)
No. of atoms = No. of moles x Avogadro No.
=> N = n x N(0)
Since, n = m/M
=> N = (m/M) x N(0)
=> N = (54/27) x 6.022 x 10^23
=> N = 2 x 6.022 x 10^23
=> N = 12.044x 10^23
=> N = 120.44 x 10^22
Therefore,
4) 54g of Aluminum has highest number of atoms.
✔️ Verified Answer
4) 54g of Aluminum has highest number of atoms.
Explanation:
(1) 10.8g of Ag (108)
N = n × N(A)
= (Wt/M) x N(A)
= (10.8/108) x 6.022 x 10²³
= 0.1 x 6.022 x 10²³
= 6.022 x 10²²
(2) 5.6g of Fe (56)
N = (Wt/M) x N(A)
= (5.6/56) x 6.022 x 10²³
= 0.1 x 6.022 x 10²³
= 6.022 x 10²²
(3) 2.4g of C (12)
N = (Wt/M) x N(A)
= (2.4/12) x 6.022 x 10²³
= 0.2 x 6.022 x 10²³
= 1.2044 x 10²³
= 12.044 x 10²²
(4) 54g of Al (27)
N = (Wt/M) x N(A)
= (54/27) x 6.022 x 10²³
= 2 x 6.022 x 10²³
= 12.044x 10²³
= 120.44 x 10²²
So, it's clearly visible that (4) Al has maximum number of atoms.