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Which of the following compound contains...

Which of the following compound contains maximum number of oxygen atoms ?

A

`4.4` gram `CO_(2)`

B

`1` gram `C_(6)H_(12)O_(6)`

C

`2` gram `O_(2)`

D

`10` gram `H_(2)C_(2)O_(4)`

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To determine which compound contains the maximum number of oxygen atoms, we will calculate the number of oxygen atoms in each option using the formula: \[ \text{Number of Oxygen Atoms} = \left( \frac{\text{mass of substance}}{\text{molar mass of substance}} \right) \times N_A \times \text{number of oxygen atoms in the formula} \] Where \(N_A\) is Avogadro's number. ### Step-by-Step Solution: **Option A: 4.4 g of CO2** 1. Calculate moles of CO2: \[ \text{Moles of CO2} = \frac{4.4 \text{ g}}{44 \text{ g/mol}} = 0.1 \text{ moles} \] 2. Number of oxygen atoms in CO2 = 2. 3. Total number of oxygen atoms: \[ \text{Total O Atoms} = 0.1 \times N_A \times 2 = 0.2 N_A \] **Option B: 1 g of C6H12O6 (Glucose)** 1. Calculate moles of C6H12O6: \[ \text{Moles of C6H12O6} = \frac{1 \text{ g}}{180 \text{ g/mol}} \approx 0.00556 \text{ moles} \] 2. Number of oxygen atoms in C6H12O6 = 6. 3. Total number of oxygen atoms: \[ \text{Total O Atoms} = 0.00556 \times N_A \times 6 \approx 0.0333 N_A \] **Option C: 2 g of O2** 1. Calculate moles of O2: \[ \text{Moles of O2} = \frac{2 \text{ g}}{32 \text{ g/mol}} = 0.0625 \text{ moles} \] 2. Number of oxygen atoms in O2 = 2. 3. Total number of oxygen atoms: \[ \text{Total O Atoms} = 0.0625 \times N_A \times 2 = 0.125 N_A \] **Option D: 10 g of H2C2O4 (Oxalic Acid)** 1. Calculate moles of H2C2O4: \[ \text{Moles of H2C2O4} = \frac{10 \text{ g}}{90 \text{ g/mol}} \approx 0.1111 \text{ moles} \] 2. Number of oxygen atoms in H2C2O4 = 4. 3. Total number of oxygen atoms: \[ \text{Total O Atoms} = 0.1111 \times N_A \times 4 \approx 0.4444 N_A \] ### Summary of Results: - Option A: \(0.2 N_A\) - Option B: \(0.0333 N_A\) - Option C: \(0.125 N_A\) - Option D: \(0.4444 N_A\) ### Conclusion: The compound with the maximum number of oxygen atoms is **Option D: 10 g of H2C2O4**, which contains approximately \(0.4444 N_A\) oxygen atoms. ---

To determine which compound contains the maximum number of oxygen atoms, we will calculate the number of oxygen atoms in each option using the formula: \[ \text{Number of Oxygen Atoms} = \left( \frac{\text{mass of substance}}{\text{molar mass of substance}} \right) \times N_A \times \text{number of oxygen atoms in the formula} \] Where \(N_A\) is Avogadro's number. ...
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