Consider the dissolution of \(\mathrm{CaCl}_{2}\) : $$\mathrm{CaCl}_{2}(s) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad \Delta H=-81.5 \mathrm{kJ}$$ An \(11.0-\mathrm{g}\) sample of \(\mathrm{CaCl}_{2}\) is dissolved in 125 g water, with both substances at \(25.0^{\circ} \mathrm{C}\). Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of \(4.18 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\).

First, we need to figure out how many moles of \(\mathrm{CaCl}_{2}\) are present in 11.0 g of the substance. To do this, we'll need the molar mass of \(\mathrm{CaCl}_{2}\): Molar mass of \(\mathrm{CaCl}_{2} = 40.08 + 2 \times 35.45 = 110.98 \, \mathrm{g/mol}\) Now, we can calculate the moles of \(\mathrm{CaCl}_{2}\): moles of \(\mathrm{CaCl}_{2} = \frac{11.0 \, \mathrm{g}}{110.98 \, \mathrm{g/mol}} = 0.099 \, \mathrm{mol}\)

We know the enthalpy of dissolution per mole is -81.5 kJ/mol. To find the total heat released during this dissolution, we'll multiply the enthalpy by the moles of \(\mathrm{CaCl}_{2}\): Total heat released, \(q = 0.099 \, \mathrm{mol} \times (-81.5 \, \mathrm{kJ/mol}) = -8.069 \, \mathrm{kJ}\) Since 1 kJ = 1000 J, the total heat released in joules is: \(q = -8.069 \, \mathrm{kJ} \times \frac{1000 \, \mathrm{J}}{1 \, \mathrm{kJ}} = -8069 \, \mathrm{J}\)

Now, we'll use the formula for specific heat capacity, which relates the heat gained or lost by a substance to its mass, specific heat capacity, and change in temperature: \(q = m \times c \times \Delta T\) Given the specific heat capacity of the solution is 4.18 J/g°C, we have: \(-8069 \, \mathrm{J} = (11.0 \, \mathrm{g} + 125 \, \mathrm{g}) \times 4.18 \, \mathrm{J/g°C} \times \Delta T\) Solve for \(\Delta T\): \(\Delta T = \frac{-8069 \, \mathrm{J}}{(11.0+125) \, \mathrm{g} \times 4.18 \, \mathrm{J/g°C}} = -13.0 \, ^{\circ} \mathrm{C}\) Now, we just need to add the change in temperature to the initial temperature: \(T_{final} = T_{initial} + \Delta T = 25.0^{\circ} \mathrm{C} - 13.0^{\circ} \mathrm{C} = 12.0^{\circ} \mathrm{C}\)

The final temperature of the \(\mathrm{CaCl}_{2}\) solution after dissolution is 12.0°C, assuming there is no heat loss to the surroundings and the specific heat capacity of the solution is 4.18 J/g°C.