QJEEAdv03Che726: JEE Advanced 2003 Chemistry - Class 11 JEEAdv03Che726 Solution: Which has maximum number of atoms?

To determine which option has the maximum number of atoms, we will use Avogadro's number and the concept of moles. Avogadro's number ( $6.02 \times 10^{23}$ atoms/mol) represents the number of atoms in one mole of any substance. The number of moles of a substance is calculated by dividing the given mass of the substance by its molar mass (which is given in grams per mole and is numerically equal to the element's atomic mass for pure elements).

Let's calculate the number of moles for each option:

Option A: 24 g of C (Carbon, atomic mass = 12 g/mol)

The number of moles of C = $\frac{24 \text{ g}}{12 \text{ g/mol}} = 2 \text{ moles}$

Therefore, the number of atoms in 24 g of C = $2 \text{ moles} \times 6.02 \times 10^{23} \text{ atoms/mol} = 1.204 \times 10^{24}$ atoms

Option B: 56 g of Fe (Iron, atomic mass = 56 g/mol)

The number of moles of Fe = $\frac{56 \text{ g}}{56 \text{ g/mol}} = 1 \text{ mole}$

Therefore, the number of atoms in 56 g of Fe = $1 \text{ mole} \times 6.02 \times 10^{23} \text{ atoms/mol} = 6.02 \times 10^{23}$ atoms

Option C: 27 g of Al (Aluminum, atomic mass = 27 g/mol)

The number of moles of Al = $\frac{27 \text{ g}}{27 \text{ g/mol}} = 1 \text{ mole}$

Therefore, the number of atoms in 27 g of Al = $1 \text{ mole} \times 6.02 \times 10^{23} \text{ atoms/mol} = 6.02 \times 10^{23}$ atoms

Option D: 108 g of Ag (Silver, atomic mass = 108 g/mol)

The number of moles of Ag = $\frac{108 \text{ g}}{108 \text{ g/mol}} = 1 \text{ mole}$

Therefore, the number of atoms in 108 g of Ag = $1 \text{ mole} \times 6.02 \times 10^{23} \text{ atoms/mol} = 6.02 \times 10^{23}$ atoms

Comparing the number of atoms in each option, Option A with 1.204 $\times 10^{24}$ atoms has the maximum number of atoms.