A student conducts an experiment to determine the enthalpy of solution for lithium chloride dissolved in water. the student combines 5.00 grams of lithium chloride with 100.0 ml of distilled water. the initial temperature of the water is 23.0°c and the highest temperature after mixing reaches 33.0°c. assume a density of 1.00 g/ml and a specific heat of 4.18 j g • °c . the calorimeter constant was found to be 20.0 j
The Correct Answer and Explanation is:
To calculate the enthalpy of solution for lithium chloride (LiCl) dissolved in water, we need to follow these steps:
Step 1: Calculate the heat absorbed by the water.
The formula to calculate the heat absorbed or released by a substance is:
[
q = m \cdot c \cdot \Delta T
]
Where:
- (q) is the heat absorbed or released (in joules),
- (m) is the mass of the substance (in grams),
- (c) is the specific heat capacity of the substance (in J/g°C),
- (\Delta T) is the change in temperature (in °C).
The mass of the water can be found using its density and volume:
[
m_{\text{water}} = \text{density} \times \text{volume}
]
Since the density of water is 1.00 g/mL and the volume is 100.0 mL, the mass of water is:
[
m_{\text{water}} = 1.00 \, \text{g/mL} \times 100.0 \, \text{mL} = 100.0 \, \text{g}
]
The change in temperature ((\Delta T)) is:
[
\Delta T = T_{\text{final}} – T_{\text{initial}} = 33.0^\circ C – 23.0^\circ C = 10.0^\circ C
]
Now we can calculate the heat absorbed by the water:
[
q_{\text{water}} = 100.0 \, \text{g} \times 4.18 \, \text{J/g°C} \times 10.0^\circ C = 418.0 \, \text{J}
]
Step 2: Account for the calorimeter constant.
The calorimeter constant accounts for the heat absorbed by the calorimeter itself. The heat absorbed by the calorimeter is:
[
q_{\text{calorimeter}} = 20.0 \, \text{J}
]
Thus, the total heat released by the dissolution of lithium chloride is:
[
q_{\text{total}} = q_{\text{water}} + q_{\text{calorimeter}} = 418.0 \, \text{J} + 20.0 \, \text{J} = 438.0 \, \text{J}
]
Step 3: Calculate the moles of lithium chloride dissolved.
The molar mass of lithium chloride (LiCl) is:
[
M_{\text{LiCl}} = 6.94 \, \text{g/mol (Li)} + 35.45 \, \text{g/mol (Cl)} = 42.39 \, \text{g/mol}
]
The number of moles of lithium chloride is:
[
n_{\text{LiCl}} = \frac{\text{mass of LiCl}}{\text{molar mass}} = \frac{5.00 \, \text{g}}{42.39 \, \text{g/mol}} = 0.118 \, \text{mol}
]
Step 4: Calculate the enthalpy of solution.
The enthalpy of solution ((\Delta H_{\text{sol}})) is the heat released per mole of solute. We can calculate it as:
[
\Delta H_{\text{sol}} = \frac{q_{\text{total}}}{n_{\text{LiCl}}} = \frac{438.0 \, \text{J}}{0.118 \, \text{mol}} = 3.72 \, \text{kJ/mol}
]
Final Answer:
The enthalpy of solution for lithium chloride in water is (3.72 \, \text{kJ/mol}).