Question

What is the heat of hydration for MgSO4·7H2O, given that its enthalpy change per mole of hydration is -192 kJ/mol and 10 grams of MgSO4·7H2O is dissolved in 100 grams of water at 25°C?

Answer 1

To calculate the heat of hydration for MgSO47H2O, we first need to determine the number of moles of MgSO47H2O in the 10 grams sample.The molar mass of MgSO47H2O is:Mg = 24.31 g/molS = 32.07 g/molO = 16.00 g/mol  x 4 = 64.00 g/mol H = 1.01 g/mol  x 14 = 14.14 g/mol Total molar mass = 24.31 + 32.07 + 64.00 + 14.14 = 134.52 g/molNow, we can find the number of moles in the 10 grams sample:moles = mass / molar massmoles = 10 g / 134.52 g/mol = 0.0743 molNow that we have the number of moles, we can calculate the heat of hydration:Heat of hydration = moles  enthalpy change per mole of hydrationHeat of hydration = 0.0743 mol   -192 kJ/mol  = -14.266 kJSo, the heat of hydration for the 10 grams of MgSO47H2O is -14.266 kJ.