The numbe of hydrogen atoms in 0.9 gm glucose, `C_(6)H_(12)O_(6)`, is same as

Options

A. 0.048 gm hydrazine.`N_(2)H_(4)`
B. 0.17 gm ammonia, `NH_(3)`
C. 0.30 gm of ethane ,`C_(2)H_(6)`
D. 0.03 gm pentene, `H_(2)`

Answer

To find the number of hydrogen atoms in 0.9 grams of glucose (C₆H₁₂O₆) and compare it with the number of hydrogen atoms in other compounds, we can follow these steps:

### Step 1: Calculate the molar mass of glucose (C₆H₁₂O₆)
- The molar mass can be calculated as follows:
- Carbon (C): 12 g/mol × 6 = 72 g/mol
- Hydrogen (H): 1 g/mol × 12 = 12 g/mol
- Oxygen (O): 16 g/mol × 6 = 96 g/mol
- Total molar mass of glucose = 72 + 12 + 96 = **180 g/mol**

**Hint:** Remember to multiply the atomic mass of each element by the number of atoms of that element in the molecule.

### Step 2: Calculate the number of moles of glucose in 0.9 grams
- Use the formula for moles:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.9 \text{ g}}{180 \text{ g/mol}} = 0.005 \text{ moles} \]

**Hint:** The number of moles can be found by dividing the mass of the substance by its molar mass.

### Step 3: Calculate the total number of hydrogen atoms in glucose
- Each molecule of glucose contains 12 hydrogen atoms.
- Therefore, the total number of hydrogen atoms can be calculated as:
\[ \text{Number of hydrogen atoms} = 12 \times \text{number of moles of glucose} \times N_A \]
Where \( N_A \) (Avogadro's number) is approximately \( 6.022 \times 10^{23} \) particles/mol.
- Substituting the values:
\[ \text{Number of hydrogen atoms} = 12 \times 0.005 \times N_A = 0.06 N_A \]

**Hint:** To find the total number of atoms, multiply the number of moles by Avogadro's number and the number of atoms per molecule.

### Step 4: Compare with other options
Now we will compare \( 0.06 N_A \) with the number of hydrogen atoms in the other compounds given in the options.

1. **For 0.048 g of hydrazine (N₂H₄)**:
- Molar mass of N₂H₄ = 32 g/mol
- Moles of hydrazine = \( \frac{0.048}{32} = 0.0015 \)
- Hydrogen atoms = \( 4 \times 0.0015 \times N_A = 0.006 N_A \)

2. **For 0.17 g of ammonia (NH₃)**:
- Molar mass of NH₃ = 17 g/mol
- Moles of ammonia = \( \frac{0.17}{17} = 0.01 \)
- Hydrogen atoms = \( 3 \times 0.01 \times N_A = 0.03 N_A \)

3. **For 0.3 g of ethane (C₂H₆)**:
- Molar mass of C₂H₆ = 30 g/mol
- Moles of ethane = \( \frac{0.3}{30} = 0.01 \)
- Hydrogen atoms = \( 6 \times 0.01 \times N_A = 0.06 N_A \)

4. **For 0.03 g of hydrogen (H₂)**:
- Molar mass of H₂ = 2 g/mol
- Moles of hydrogen = \( \frac{0.03}{2} = 0.015 \)
- Hydrogen atoms = \( 2 \times 0.015 \times N_A = 0.03 N_A \)

### Conclusion
The only option that matches the number of hydrogen atoms in 0.9 g of glucose (which is \( 0.06 N_A \)) is **0.3 g of ethane (C₂H₆)**.

**Final Answer:** The number of hydrogen atoms in 0.9 g of glucose is the same as in **0.3 g of ethane (C₂H₆)**.