No. of molecules in 88 gm CO2 is y × 1023 , then value of y is

Key Ideas Needed to Crack the Problem

1. Molar Mass ($M$)
   • For CO$_2$, add the atomic masses: $M = 12\,(\text{for C}) + 2\times16\,(\text{for O}) = 44\,\text{g mol}^{-1}$.

2. Number of Moles ($n$)
   $n = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{m}{M}$

3. Avogadro’s Number ($N_A$)
   One mole contains $N_A = 6.022\times10^{23}$ particles (atoms, molecules, ions …).

4. Total Particles ($N$)
   $N = n\,N_A$

5. Form of the Answer
   The question wants $N$ expressed as $y\times10^{23}$, so we only need the coefficient $y$.

Why Each Step Matters

• Finding $M$ tells you how much one mole weighs.
• Finding $n$ converts the mass you have into moles.
• Multiplying by $N_A$ translates moles into actual count of molecules.
• Arranging into $y\times10^{23}$ fits the format the examiner demands.

🏫 Imagine counting biscuits in big boxes

1. One Big Box (1 mole) of biscuits has $6.022\times10^{23}$ biscuits inside ‒ that’s called Avogadro’s number.

2. A packet of CO$_2$ gas that weighs 44 g is exactly one big box (1 mole).

3. Our packet is 88 g, so that’s two big boxes (2 moles).

4. Two boxes have double the biscuits:

   $2 \times 6.022\times10^{23} \approx 12.04\times10^{23}$

5. The question writes this as $y\times10^{23}$, so $y \approx 12.04$ (usually rounded to 12).

Hence, $y=12$ (or 12.04 if your teacher wants more digits).

Step-by-Step Solution

1. Molar mass of CO$_2$
   $M = 12 + 2\times16 = 44\,\text{g mol}^{-1}$

2. Number of moles in 88 g
   $n = \frac{m}{M} = \frac{88\,\text{g}}{44\,\text{g mol}^{-1}} = 2\,\text{mol}$

3. Total molecules
   $N = nN_A = 2\times6.022\times10^{23} = 12.044\times10^{23}$

4. Match the required form
   $N = y\times10^{23}\;\Rightarrow\;y\approx12.044$

5. Rounded value
   Most exams round to two significant figures: $y=12$.

Final Answer: $\boxed{y\approx12}$