When 0.250 moles of LiCl are added to 200.0 g of water in a constant pressure calorimeter a temperature change of +11.08 °C is observed. Given that the specific heat of the resulting solution is 4.184 J g^-1 °C and we can ignore the small amount of energy absorbed by the calorimeter, what is the molar enthalpy of solution ($\Delta$H_sol)for LiCl?Hint: Do not ignore the specific heat of water in this problem.

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Quizplus · Accepted Answer

The heat absorbed by the solution is:$$q = mC_p \Delta T = (200.0 g)(4.184 J g^{-1} °C)(11.08 °C) = 9206 J$$The molar enthalpy of solution is:$$\Delta H_{sol} = \frac{q}{n} = \frac{9206 J}{0.250 mol} = -37.1 kJ/mol$$The negative sign indicates that the process is exothermic.

When 0250 Moles of LiCl Are Added to 200 Δ\DeltaΔ Hsol)for LiCl?Hint: Do Not Ignore the Specific Heat of Water

When 0250 Moles of LiCl Are Added to 200 $\Delta$ H_sol)for LiCl?Hint: Do Not Ignore the Specific Heat of Water