a 4.50 sample of LiCl at 25.0*C dissolves in 25.0 mL of water also at 25.0*C. The final equilibrium temperature of resulting solution is 60.8*C. What is the enthalpy of solution, of LiCL expressed in kilojoules per mole?
To determine the enthalpy of solution for LiCl, we can use the equation:
ΔH = q / n
Where:
- ΔH is the enthalpy of solution
- q is the heat exchanged during the solution process
- n is the number of moles of solute (LiCl)
To calculate the heat exchanged during the solution process, we can use the formula:
q = m * Cp * ΔT
Where:
- q is the heat exchanged
- m is the mass of the solution
- Cp is the specific heat capacity of water
- ΔT is the change in temperature
First, we need to calculate the mass of LiCl used. Given that the sample is 4.50 g and the molar mass of LiCl is 42.39 g/mol, we can find the number of moles (n):
n = 4.50 g / 42.39 g/mol
Next, for the heat exchanged, we need to calculate the change in temperature (ΔT). The starting temperature is 25.0°C, and the final temperature is 60.8°C. So:
ΔT = 60.8°C - 25.0°C
Now, we need to calculate the heat capacity of the water (Cp). The specific heat capacity of water is approximately 4.184 J/g°C. However, we need to convert the mass of water from milliliters (mL) to grams (g) since the specific heat capacity is in J/g°C. The density of water is approximately 1 g/mL. Therefore:
m_water = 25.0 mL * 1 g/mL
With the mass of water, we can calculate Cp:
Cp = 4.184 J/g°C
Now that we have all the necessary values, we can substitute them into the equation:
q = m * Cp * ΔT
Let's calculate the values:
n = 4.50 g / 42.39 g/mol = 0.1063 mol
ΔT = 60.8°C - 25.0°C = 35.8°C
m_water = 25.0 mL * 1 g/mL = 25.0 g
Cp = 4.184 J/g°C
Now we substitute the values:
q = 25.0 g * 4.184 J/g°C * 35.8°C
Now we can calculate q.