Answer
$26\text{ kJ/mol}$
Work Step by Step
$q=mc\Delta T$
$q=75.0*4.18*(23.34-25.00)=-520.41\text{ J}$
$1.60\text{ grams $NH_4NO_3$}*\frac{1\text{ mol $NH_4NO_3$}}{80.043\text{ grams $NH_4NO_3$}}=0.0200$
We switch the signs of the enthalpy change as the system is the $NH_4OH$ which is absorbing energy.
$\frac{520.14\text{ J}}{0.200\text{ mols}}=26\text{ kJ/mol}$
