Number of atoms present in 1.8g `H_(2)O, 1.7g NH_(3)` and `1.6g CH_(4)` has the following sequence

Options

A. `H_(2)O lt NH_(3) lt CH_(4)`
B. `CH_(4) lt NH_(3) lt H_(2)O`
C. `CH_(4)= NH_(3)= H_(2)O`
D. `NH_(3) lt CH_(4) lt H_(2)O`

Explanation

To find the number of atoms present in 1.8 g of \( H_2O \), 1.7 g of \( NH_3 \), and 1.6 g of \( CH_4 \), we will use the formula:

\[ \text{Number of atoms} = \left( \frac{\text{Given weight}}{\text{Molecular weight}} \right) \times N_A \times \text{Number of atoms in one mole} \]

where \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \) atoms/mole).

### Step 1: Calculate the number of atoms in \( H_2O \)

1. **Given weight**: 1.8 g
2. **Molecular weight of \( H_2O \)**: \( 2(1) + 16 = 18 \) g/mole
3. **Number of atoms in one mole of \( H_2O \)**: 3 (2 H + 1 O)

Using the formula:

\[ \text{Number of atoms in } H_2O = \left( \frac{1.8 \, \text{g}}{18 \, \text{g/mole}} \right) \times N_A \times 3 \]

Calculating this:

\[ = \left( \frac{1.8}{18} \right) \times 6.022 \times 10^{23} \times 3 \]
\[ = 0.1 \times 6.022 \times 10^{23} \times 3 \]
\[ = 0.3 \times 6.022 \times 10^{23} = 0.3 N_A \]

### Step 2: Calculate the number of atoms in \( NH_3 \)

1. **Given weight**: 1.7 g
2. **Molecular weight of \( NH_3 \)**: \( 14 + 3(1) = 17 \) g/mole
3. **Number of atoms in one mole of \( NH_3 \)**: 4 (1 N + 3 H)

Using the formula:

\[ \text{Number of atoms in } NH_3 = \left( \frac{1.7 \, \text{g}}{17 \, \text{g/mole}} \right) \times N_A \times 4 \]

Calculating this:

\[ = \left( \frac{1.7}{17} \right) \times 6.022 \times 10^{23} \times 4 \]
\[ = 0.1 \times 6.022 \times 10^{23} \times 4 \]
\[ = 0.4 \times 6.022 \times 10^{23} = 0.4 N_A \]

### Step 3: Calculate the number of atoms in \( CH_4 \)

1. **Given weight**: 1.6 g
2. **Molecular weight of \( CH_4 \)**: \( 12 + 4(1) = 16 \) g/mole
3. **Number of atoms in one mole of \( CH_4 \)**: 5 (1 C + 4 H)

Using the formula:

\[ \text{Number of atoms in } CH_4 = \left( \frac{1.6 \, \text{g}}{16 \, \text{g/mole}} \right) \times N_A \times 5 \]

Calculating this:

\[ = \left( \frac{1.6}{16} \right) \times 6.022 \times 10^{23} \times 5 \]
\[ = 0.1 \times 6.022 \times 10^{23} \times 5 \]
\[ = 0.5 \times 6.022 \times 10^{23} = 0.5 N_A \]

### Step 4: Summary of results

- Number of atoms in \( H_2O \): \( 0.3 N_A \)
- Number of atoms in \( NH_3 \): \( 0.4 N_A \)
- Number of atoms in \( CH_4 \): \( 0.5 N_A \)

### Conclusion

The sequence of the number of atoms present in the compounds is:

\[ CH_4 > NH_3 > H_2O \]