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Calculate the heat of hydration when 10.0 g of calcium chloride (CaCl2) dissolves in 100 g of water at 25°C. The enthalpy of solution for calcium chloride is -81.5 kJ/mol.

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To calculate the heat of hydration, we first need to determine the number of moles of calcium chloride  CaCl2  that are present in the 10.0 g sample.The molar mass of CaCl2 is:Ca = 40.08 g/molCl = 35.45 g/mol  and there are 2 Cl atoms So, the molar mass of CaCl2 = 40.08 +  2  35.45  = 111.0 g/molNow, we can calculate the number of moles of CaCl2 in the 10.0 g sample:moles of CaCl2 = mass / molar mass = 10.0 g / 111.0 g/mol = 0.09009 molThe enthalpy of solution for calcium chloride is given as -81.5 kJ/mol. This means that when 1 mole of CaCl2 dissolves in water, it releases 81.5 kJ of heat. We can use this information to calculate the heat of hydration for the 0.09009 mol of CaCl2:Heat of hydration = moles of CaCl2  enthalpy of solution per moleHeat of hydration = 0.09009 mol  -81.5 kJ/mol = -7.34 kJSo, the heat of hydration when 10.0 g of calcium chloride dissolves in 100 g of water at 25C is -7.34 kJ.

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