a Coffee cup calorimeter initially contains 125 g of water , at a temperature of `24.2^(@)`C ,8 of ammonium nitrate `(NH_(4)NO_(3))` also at `24.2^(@)C`is added to the Water and the final temperature is `18.2^(@)c` What is the heat of solution of ammonium nitrate in KJ /mol? (The specific heat capacity of the solution is `4.2J//^(@)c`.)

Options

A. 33.51 kJ/mol
B. 39.5 kJ/mol
C. 32.2 kJ/mol
D. 37.3 kJ/mol

Explanation

To find the heat of solution of ammonium nitrate (NH₄NO₃) in kilojoules per mole, we can follow these steps:

### Step 1: Calculate the change in temperature (ΔT)
The initial temperature (T_initial) is 24.2°C and the final temperature (T_final) is 18.2°C.

\[ \Delta T = T_{\text{initial}} - T_{\text{final}} = 24.2°C - 18.2°C = 6.0°C \]

### Step 2: Calculate the total mass of the solution
The mass of water is 125 g and the mass of ammonium nitrate added is 8 g.

\[ \text{Total mass} = \text{mass of water} + \text{mass of NH}_4\text{NO}_3 = 125 \, \text{g} + 8 \, \text{g} = 133 \, \text{g} \]

### Step 3: Calculate the heat absorbed by the solution (q)
Using the formula for heat:

\[ q = m \cdot C_p \cdot \Delta T \]

Where:
- \(m\) = mass of the solution (133 g)
- \(C_p\) = specific heat capacity of the solution (4.2 J/°C)
- \(\Delta T\) = change in temperature (6.0°C)

Substituting the values:

\[ q = 133 \, \text{g} \cdot 4.2 \, \text{J/°C} \cdot 6.0 \, \text{°C} \]

Calculating \(q\):

\[ q = 133 \cdot 4.2 \cdot 6.0 = 3353.6 \, \text{J} \]

### Step 4: Convert heat from joules to kilojoules
To convert joules to kilojoules, divide by 1000:

\[ q = \frac{3353.6 \, \text{J}}{1000} = 3.3536 \, \text{kJ} \]

### Step 5: Calculate the number of moles of ammonium nitrate
The molecular weight of ammonium nitrate (NH₄NO₃) is approximately 80 g/mol.

Using the formula for moles:

\[ \text{Moles of NH}_4\text{NO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{8 \, \text{g}}{80 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 6: Calculate the heat of solution per mole
The heat of solution (ΔH) is defined as the heat absorbed per mole of solute:

\[ \Delta H = \frac{q}{\text{moles}} = \frac{3.3536 \, \text{kJ}}{0.1 \, \text{mol}} = 33.536 \, \text{kJ/mol} \]

### Step 7: Round to appropriate significant figures
Rounding to three significant figures, we find:

\[ \Delta H \approx 33.5 \, \text{kJ/mol} \]

### Final Answer
The heat of solution of ammonium nitrate is approximately **33.5 kJ/mol**.

---