The vapour density of fully dissociated `NH_(4)Cl` would be

Options

A. Less than half of the vapour density of pure `NH_(4)Cl`
B. Double of the vapour density of pure `NH_(4)Cl`
C. Half of the vapour density of pure `NH_(4)Cl`
D. One-third of the vapour density of pure `NH_(4)Cl`

Answer

To find the vapor density of fully dissociated \( NH_4Cl \), we can follow these steps:

### Step 1: Understand the dissociation of \( NH_4Cl \)
When \( NH_4Cl \) fully dissociates, it breaks down into two gaseous products:
\[ NH_4Cl \rightarrow NH_3 + HCl \]
This means that one mole of \( NH_4Cl \) produces one mole of \( NH_3 \) and one mole of \( HCl \).

### Step 2: Determine the initial conditions
Initially, we have one mole of \( NH_4Cl \) in a container. The vapor density (VD) of a substance is defined as:
\[ VD = \frac{\text{Molar mass}}{2} \]
For \( NH_4Cl \), the molar mass can be calculated as follows:
- Molar mass of \( N \) = 14 g/mol
- Molar mass of \( H \) = 1 g/mol (4 H in \( NH_4 \))
- Molar mass of \( Cl \) = 35.5 g/mol

Thus, the molar mass of \( NH_4Cl \) is:
\[ 14 + (4 \times 1) + 35.5 = 14 + 4 + 35.5 = 53.5 \text{ g/mol} \]
The vapor density of pure \( NH_4Cl \) is:
\[ VD_{NH_4Cl} = \frac{53.5}{2} = 26.75 \text{ g/L} \]

### Step 3: Analyze the dissociation effect on vapor density
After dissociation, we have 2 moles of gas (1 mole of \( NH_3 \) and 1 mole of \( HCl \)) in the same volume. According to the ideal gas law, if the number of moles of gas increases while the volume remains constant, the pressure will increase, but the vapor density will be affected by the number of moles.

### Step 4: Calculate the new vapor density
Since the number of moles has doubled (from 1 mole to 2 moles), the vapor density will be inversely proportional to the number of moles. Therefore, the new vapor density after dissociation will be:
\[ VD_{dissociated} = \frac{VD_{NH_4Cl}}{2} = \frac{26.75}{2} = 13.375 \text{ g/L} \]

### Conclusion
Thus, the vapor density of fully dissociated \( NH_4Cl \) is half of the vapor density of pure \( NH_4Cl \).

The correct answer is:
**Half of the vapor density of pure \( NH_4Cl \)**.

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