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# $100\; mL$ of $0.10M\; HCl$ is mixed with $100\; mL$ of $0.10M\; NaOH.$ The solution temperature rises by $6.0 ^{\circ}C.$ Calculate the enthalpy of neutralization per mole of $HCl.$

$(a)\;-5.02 \times 102\;kJ \\(b)\; -1.3 \times 102\;kJ \\(c)\;-8.4 \times 101 \;kJ \\(d)\;-6.3 \times 101\;kJ$

The enthalpy of neutralization per mole of $HCl.$ is : $-5.02 \times 102\;kJ$
Hence a is the correct answer.