A sample of 2.50 g LiI is added to 75.0 mL of water initially at 21.0°C. After stirring and dissolving, the temperature of the solution is 24.3°C. Assume that the specific heat of the solution is 4.184 J/g°C, the density is 1.00 g/mL, and the calorimeter constant is 40.0 J/°C. Determine the heat of solution for LiI (ΔHsoln).
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To determine the heat of solution for LiI (ΔHsoln), we can use the equation:
q = m × c × ΔT
Where:
q is the heat absorbed or released by the solution in joules (J),
m is the mass of the solution in grams (g),
c is the specific heat capacity of the solution in J/g°C, and
ΔT is the change in temperature of the solution in °C.
First, we need to calculate the mass of the solution. The mass of the solution can be found by adding the mass of the LiI and the mass of the water:
Mass of LiI = 2.50 g
Volume of water = 75.0 mL
Density of water = 1.00 g/mL
Mass of water = Volume of water × Density of water
= 75.0 mL × 1.00 g/mL
= 75.0 g
Mass of solution = Mass of LiI + Mass of water
= 2.50 g + 75.0 g
= 77.5 g
Next, we calculate the change in temperature (ΔT):
ΔT = Final temperature - Initial temperature
= 24.3°C - 21.0°C
= 3.3°C
Now, we can calculate the heat of solution (q) using the equation:
q = m × c × ΔT
= 77.5 g × 4.184 J/g°C × 3.3°C
= 1077.2856 J
However, we need to account for the calorimeter constant (the heat absorbed or released by the calorimeter). The calorimeter constant is given as 40.0 J/°C.
q_actual = q - calorimeter constant
= 1077.2856 J - 40.0 J/°C
= 1037.2856 J
Therefore, the heat of solution for LiI (ΔHsoln) is approximately 1037.2856 J.