55.90 kJ/mol x # mol = 6.21 kJ. Solve for # mols.
Then M = #mols/L. You know # mols and L (0.075 L). Calculate M.
Given that the enthalpy of neutralization for the reaction of HCl (a strong acid) and NaOH (a strong base) is always -55.90 kJ per mole of H2O formed, what is the concentration of a 75 mL sample of HCl if the enthalpy of neutralization for the reaction was found to be -6.21 kJ, assuming no heat is lost to the calorimeter?
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