Question

which of the following elements contains the greatest no. of atoms
(a) 4g of He (b) 46g of Na
(c)0.40 g Ca (d) 12g of He

Answer 1

Answer:

iv) 12g He contains more no.of atoms.

No.of moles = No.of given atoms/avagadro

no.of atoms

No.of atoms = No.of moles × avagadro no.

(i) no.of atoms = 4g/4g × No = 1No

(ii) no.of atoms = 46g/23g × No = 2× No

(iii) no.of atoms = 0.40/40 ×No = 0.1 ×No

(iv) no.of atoms = 12g/4g × No = 3×No

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Answer 2

Answer:

(d) 12g of He

Explanation:

For comparing number of atoms, first we calculate the moles as all are monoatomic and hence, moles ×NA= number of atoms .

Moles of 4g He

=4/4

=1 mol

Moles of 46g Na

=46/23

=2 mol

Moles of 0.40g Ca

=0.40/40

=0.01 mol

Moles of 12g He

=12/4

=3 mol

Hence 12 g He contains greatest number of atoms as it possesses maximum number of moles.