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Calculate the enthalpy change of dissolving 10.0 g of sodium hydroxide (NaOH) in 200.0 mL of water at 25°C. Given: the enthalpy of dissolution of NaOH is -44.5 kJ/mol.

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To calculate the enthalpy change of dissolving 10.0 g of sodium hydroxide  NaOH  in 200.0 mL of water at 25C, we need to first determine the number of moles of NaOH being dissolved. The molar mass of NaOH is:Na = 22.99 g/molO = 16.00 g/molH = 1.01 g/molTotal = 22.99 + 16.00 + 1.01 = 40.00 g/molNow, we can calculate the number of moles of NaOH:moles = mass / molar massmoles = 10.0 g / 40.00 g/mol = 0.25 molNext, we can use the given enthalpy of dissolution to calculate the enthalpy change:H = moles  enthalpy of dissolutionH = 0.25 mol   -44.5 kJ/mol  = -11.125 kJTherefore, the enthalpy change of dissolving 10.0 g of sodium hydroxide in 200.0 mL of water at 25C is -11.125 kJ.

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