If you were to prepare a 1.00 molal solution of CaCl2 beginning with water at 27.0 C, what would the final temperature of the solution be (in C)? Assume that the specific heats of both pure H2O and the solution are the same, 4.18 J/K*g.
The dissolution of CaCl2 in water is exothermic, with deltaH soln = -81.3 kJ/mol.
Don't we need to know how much solution we are to prepare? I'll assume we are to use 1 kg water.
1.00 m CaCl2 = 110.983 g CaCl2/kg water.
What's the density of the solution. I will assume 1.00 g/mL which probably is not a valid assumption.
81,300 J = 1,000g x 4.184 J/C*g x (Tfinal-27)
Solve for Tfinal.
Check my thinking.
150.248
To determine the final temperature of the solution, we can use the principle of conservation of energy. The heat released during the dissolution of CaCl2 can be considered as the heat gained by the water. We can use the equation:
q_water = q_CaCl2
where q_water is the heat gained by the water and q_CaCl2 is the heat released by the dissolution of CaCl2.
First, we need to calculate the heat released by the dissolution of CaCl2. Given that the deltaH soln (enthalpy change) is -81.3 kJ/mol, we can use the equation:
q_CaCl2 = deltaH soln * moles of CaCl2
Since we want to prepare a 1.00 molal solution of CaCl2, we know that the moles of CaCl2 will be equal to the moles of water in the solution. In a 1.00 molal solution, there is 1.00 mole of solute (CaCl2) for every 1.00 kg of solvent (water). Since the molecular weight of CaCl2 is 110.98 g/mol, we can calculate the amount of CaCl2 needed:
mass of CaCl2 = molality * molecular weight of water
mass of CaCl2 = 1.00 molal * 1.00 kg * (110.98 g/mol / 1000 g/kg)
mass of CaCl2 = 0.11098 kg
Therefore, the moles of CaCl2 in the solution is:
moles of CaCl2 = mass of CaCl2 / molar mass of CaCl2
moles of CaCl2 = 0.11098 kg / 110.98 g/mol
moles of CaCl2 = 0.00100 mol
Now we can substitute the values into the equation to calculate the heat released:
q_CaCl2 = -81.3 kJ/mol * 0.00100 mol
q_CaCl2 = -0.0813 kJ
Now we can equate the heat gained by the water to the heat released by the dissolution of CaCl2:
q_water = q_CaCl2
Since we know that the specific heat capacity of both pure H2O and the solution is 4.18 J/K*g, we can use the equation:
q_water = m_water * C_water * deltaT
where m_water is the mass of water, C_water is the specific heat capacity of water, and deltaT is the change in temperature of the water.
Assuming the initial temperature of the water is 27.0°C and the final temperature is T°C, we can rewrite the equation as:
q_CaCl2 = m_water * C_water * (T - 27.0)
Now we can solve for T:
T - 27.0 = q_CaCl2 / (m_water * C_water)
T = (q_CaCl2 / (m_water * C_water)) + 27.0
Let's substitute the values to calculate the final temperature:
T = (-0.0813 kJ / (1.00 kg * 4.18 J/K*g)) + 27.0
Converting -0.0813 kJ to J:
T = (-0.0813 kJ * 1000 J/kJ / (1.00 kg * 4.18 J/K*g)) + 27.0
T = -19.43°C + 27.0
T = 7.57°C
Therefore, the final temperature of the solution would be approximately 7.57°C.