Calculate molality (m) of each ion present in the aqueous solution of `2M NH_(4)Cl` assuming 100% dissociation according to reaction
`NH_(4) Cl (aq) rarr NH_(4)^(+) (aq) + Cl^(-) (aq)`
Given: Density of solution = 3.107 gm/ml

Answer

To calculate the molality (m) of each ion present in the aqueous solution of 2M NH₄Cl, we can follow these steps:

### Step 1: Understand the Dissociation of NH₄Cl
The dissociation of ammonium chloride (NH₄Cl) in water can be represented as:
\[ \text{NH}_4\text{Cl} (aq) \rightarrow \text{NH}_4^+ (aq) + \text{Cl}^- (aq) \]
Since the dissociation is 100%, for every mole of NH₄Cl that dissolves, it produces 1 mole of NH₄⁺ and 1 mole of Cl⁻.

### Step 2: Determine Moles of Each Ion
Given that the solution is 2M (2 moles per liter), we can conclude that:
- In 1 liter of solution, there are 2 moles of NH₄Cl.
- Therefore, there will also be 2 moles of NH₄⁺ and 2 moles of Cl⁻.

### Step 3: Calculate the Mass of the Solution
The density of the solution is given as 3.107 g/mL. To find the mass of 1 liter (1000 mL) of the solution:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 3.107 \, \text{g/mL} \times 1000 \, \text{mL} = 3107 \, \text{g} = 3.107 \, \text{kg} \]

### Step 4: Calculate the Molar Mass of NH₄Cl
To find the molar mass of NH₄Cl:
- Molar mass of N (Nitrogen) = 14 g/mol
- Molar mass of H (Hydrogen) = 1 g/mol (4 H atoms = 4 g/mol)
- Molar mass of Cl (Chlorine) = 35.5 g/mol

Calculating the total:
\[ \text{Molar mass of NH}_4\text{Cl} = 14 + 4 + 35.5 = 53.5 \, \text{g/mol} \]

### Step 5: Calculate the Mass of Solute (NH₄Cl)
Since we have 2 moles of NH₄Cl in 1 liter of solution:
\[ \text{Mass of NH}_4\text{Cl} = \text{Number of moles} \times \text{Molar mass} = 2 \, \text{moles} \times 53.5 \, \text{g/mol} = 107 \, \text{g} = 0.107 \, \text{kg} \]

### Step 6: Calculate the Mass of the Solvent
The mass of the solvent (water) can be calculated by subtracting the mass of the solute from the mass of the solution:
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} = 3.107 \, \text{kg} - 0.107 \, \text{kg} = 3.000 \, \text{kg} \]

### Step 7: Calculate the Molality of Each Ion
The formula for molality (m) is:
\[ m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \]

For each ion (NH₄⁺ and Cl⁻):
\[ m = \frac{2 \, \text{moles}}{3.000 \, \text{kg}} = 0.667 \, \text{mol/kg} \]

### Final Answer
The molality of each ion (NH₄⁺ and Cl⁻) in the solution is approximately:
\[ \text{Molality} = 0.667 \, \text{mol/kg} \]

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