When a 8.00-g sample of KBr is dissolved in water in a calorimeter that has a total heat capacity of 3.344 kJ·K–1, the temperature decreases by 0.400 K. Calculate the molar heat of solution of KBr.
To calculate the molar heat of solution (ΔHsoln) of KBr, we need to use the equation:
ΔHsoln = q / n
where ΔHsoln is the molar heat of solution, q is the heat transferred to the solution, and n is the number of moles of solute (KBr).
First, let's calculate the heat transferred to the solution (q):
q = CΔT
where C is the heat capacity of the calorimeter and ΔT is the change in temperature of the solution.
Given:
C = 3.344 kJ·K–1 (heat capacity of the calorimeter)
ΔT = -0.400 K (temperature decrease)
q = (3.344 kJ·K–1) * (-0.400 K)
q = -1.338 kJ
Next, we need to calculate the number of moles of KBr (n) in the 8.00 g sample. To do this, we need to know the molar mass of KBr. The molar mass of KBr can be calculated by adding the atomic masses of potassium (K) and bromine (Br) from the periodic table.
Potassium (K) has a molar mass of 39.10 g/mol.
Bromine (Br) has a molar mass of 79.90 g/mol.
Molar mass of KBr = 39.10 g/mol + 79.90 g/mol
Molar mass of KBr = 119.00 g/mol
Now, we can calculate the number of moles (n) of KBr:
n = mass / molar mass
Given:
mass = 8.00 g (sample mass)
n = (8.00 g) / (119.00 g/mol)
n = 0.0672 mol
Finally, we can calculate the molar heat of solution (ΔHsoln) using the equation:
ΔHsoln = q / n
ΔHsoln = (-1.338 kJ) / (0.0672 mol)
ΔHsoln = -19.93 kJ/mol
Therefore, the molar heat of solution of KBr is approximately -19.93 kJ/mol.