Question

3. 2.0 g of oxygen contains number of atoms equal to that
in
4.0 g of sulphur (b) 7.0 g of nitrogen
(e) 0.5 g of hydrogen (d) 2.3 g of sodium.
(JIPMER)​

Answer 1

Question:-

2.0 g of oxygen contains number of atoms equal to that in:-

a) 4.0 g of sulphur

b) 7.0 g of Nitrogen

c) 0.5 g of hydrogen

d) 2.3 g of sodium

Answer :-

Option A.

4.0 g of Sulphur.

Explanation:-

16 g of oxygen contain $N_a$ atoms

2.0 g of oxygen contain = $\frac {N_a}{16} × 2$ atoms

2.0 g of oxygen contain = $\frac{N_a}{8}$ atoms

Now,let check which option is correct.

1) 32 g of Sulphur has $N_a$atoms

4.0 g of sulphur has $\frac{N_a}{32}× 4$atoms

4.0 g of sulphur has $\frac{N_a}{8}$atoms

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2) 14 g of Nitrogen has $N_a$atoms

then, 7 g of Nitrogen has $\frac{N_a}{14}×7$atoms

7 g of Nitrogen has $\frac{N_a}{2}$atoms

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3) 1 g of hydrogen has $N_a$atoms

then, 0.5 g of hydrogen has $\frac{N_a}{2}$atoms

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4) 21 g of sodium has $N_a$atoms

$\therefore$ 2.1 g of Sodium has $\frac{N_a}{10}$atoms

From above solution ,

The correct answer is 4.0 g of sulphur has same number of atoms as 2.0 g of Oxygen.

where $N_a$ is $6.022×10^{23} atoms$or Avagardo constant