No. of molecules & atoms in 54 gm H2O is y × 1023 and z × 1023 respectively then value of y & z is (NA = 6.022 × 1023 )

Key Concepts Needed

1. Mole: One mole of any substance contains Avogadro’s number ($N_A$) of elementary particles.
2. Molar mass of water: $H_2O$ has mass $2\times1 + 16 = 18\,\text{g mol}^{-1}$.
3. Conversion steps:
   • Convert given mass to moles using $\text{moles} = \frac{\text{given mass}}{\text{molar mass}}$.
   • Multiply moles by $N_A$ to get number of molecules.
   • Multiply molecules by the number of atoms per molecule to get total atoms.

Logical Chain a Student Should Follow

1. Find moles: $\text{n} = \frac{54\,\text{g}}{18\,\text{g mol}^{-1}} = 3\,\text{mol}$

2. Find molecules:

   $\text{molecules} = n \times N_A = 3 \times 6.022 \times 10^{23} = 18.066 \times 10^{23}$

3. Find atoms: Each $H_2O$ has 3 atoms.

   $\text{atoms} = 3 \times (18.066 \times 10^{23}) = 54.198 \times 10^{23}$

4. Identify y and z: Comparing with $y \times 10^{23}$ and $z \times 10^{23}$ gives $y \approx 18.066,\qquad z \approx 54.198$

Why each step?

• Step-1 links mass with amount (moles).
• Step-2 connects amount with count (molecules) via $N_A$.
• Step-3 uses molecular formula to scale up from molecules to atoms.

Imagine sharing candies

1. Each candy box is like 1 water molecule (H₂O).
2. One big bag always holds 6.022 ×10²³ candy boxes. Scientists call that gigantic number Avogadro’s number (N_A).
3. Weight of one candy box (one H₂O molecule) is such that 18 g of water makes exactly 1 big bag (1 mole).
4. We are given 54 g of water. Think of it as 54 kg of candies if each box weighed 18 kg.
5. How many big bags? 54 kg ÷ 18 kg per bag = 3 bags (3 moles).
6. A bag contains 6.022 × 10²³ boxes, so 3 bags contain 3 × 6.022 × 10²³ boxes = 18.066 × 10²³ water molecules.
7. Each box has 3 tiny pieces (2 H atoms + 1 O atom). So multiply by 3 again: 54.198 × 10²³ atoms in total.

So the two magic numbers are:

• y ≈ 18.066 (for molecules)
• z ≈ 54.198 (for atoms)

Step-by-Step Calculation

1. Determine moles of water

   $n = \frac{m}{M} = \frac{54\,\text{g}}{18\,\text{g mol}^{-1}} = 3\,\text{mol}$

2. Calculate number of molecules

   $\text{molecules} = n \times N_A = 3 \times 6.022 \times 10^{23} = 18.066 \times 10^{23}$

   Hence $y = 18.066$.

3. Calculate total number of atoms (each $H_2O$ contains 3 atoms)

   $\text{atoms} = 3 \times 18.066 \times 10^{23} = 54.198 \times 10^{23}$

   Hence $z = 54.198$.

4. Final Answers

   $y \approx 18.07,\; z \approx 54.20$

   (values kept to four significant digits; any close rounded value is acceptable).