Which among the following has the maximum number of molecules?

Question

Accepted Answer

1. **Identify the Principle:** The number of molecules is directly proportional to the number of moles ($n$). The formula to calculate moles is:
   $$ n = \frac{	ext{Given Mass (g)}}{	ext{Molar Mass (g/mol)}} $$
   [Class 12 Chemistry, Ch 1, Example 1.1]
2. **Calculate moles for each option:**
   - **Option A ($7	ext{ g N}_2$):** Molar mass of $N_2 = 2 	imes 14 = 28	ext{ g/mol}$.
     $$ n = \frac{7}{28} = 0.25	ext{ mol} $$
   - **Option B ($2	ext{ g H}_2$):** Molar mass of $H_2 = 2 	imes 1 = 2	ext{ g/mol}$.
     $$ n = \frac{2}{2} = 1.0	ext{ mol} $$
   - **Option C ($16	ext{ g NO}_2$):** Molar mass of $NO_2 = 14 + (2 	imes 16) = 46	ext{ g/mol}$.
     $$ n = \frac{16}{46} \approx 0.35	ext{ mol} $$
   - **Option D ($16	ext{ g O}_2$):** Molar mass of $O_2 = 2 	imes 16 = 32	ext{ g/mol}$.
     $$ n = \frac{16}{32} = 0.5	ext{ mol} $$
3. **Conclusion:** Option B ($2	ext{ g H}_2$) has the highest number of moles ($1.0	ext{ mol}$), and thus the maximum number of molecules.

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