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NCERT Exemplar · Q5

Q.The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following element contains the greatest number of atoms?

(i) 4 g He
(ii) 46 g Na
(iii) 0.40 g Ca
(iv) 12 g He
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Convert each mass to moles using n=massmolar massn = \frac{\text{mass}}{\text{molar mass}}, then multiply by Avogadro's number to find atom count. The largest number of moles gives the greatest number of atoms: 12 g He contains 3 moles and thus 3NA3N_A atoms.

The question tests whether you can connect mass, molar mass, and the mole concept. Avogadro's number NA≈6.022×1023N_A \approx 6.022 \times 10^{23} tells us how many particles (atoms, in this case) are in one mole. Since every element here exists as individual atoms (not molecules), the number of atoms is simply the number of moles multiplied by NAN_A.

The strategy is straightforward: calculate the number of moles for each option, then compare. Whichever has the most moles automatically has the most atoms.

n=mass (g)molar mass (g/mol)n = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}

Now let's work through each option systematically.

1. Option (i): 4 g He

Helium has a molar mass of 4 g/mol (atomic mass ≈ 4 u).

nHe=44=1 moln_{\text{He}} = \frac{4}{4} = 1 \text{ mol}

Number of atoms = 1×NA=NA1 \times N_A = N_A

2. Option (ii): 46 g Na

Sodium has a molar mass of 23 g/mol (atomic mass ≈ 23 u).

nNa=4623=2 moln_{\text{Na}} = \frac{46}{23} = 2 \text{ mol}

Number of atoms = 2×NA=2NA2 \times N_A = 2N_A

3. Option (iii): 0.40 g Ca

Calcium has a molar mass of 40 g/mol (atomic mass ≈ 40 u).

nCa=0.4040=0.01 moln_{\text{Ca}} = \frac{0.40}{40} = 0.01 \text{ mol} …

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