Answer
$66.9\text{ kJ/mol}$
Work Step by Step
Moles of $AgNO_3 = 0.050*0.100=0.0050$
$q=mc\Delta T$
$q=100.0*4.18*(23.40-22.60)$
$q=334.4\text{ J per 0.0050 mol of $AgNO_3$ consumed}$
$\frac{334.4\text{ mol}}{0.0050\text{ mol $AgNO_3$}}=66.9\text{ kJ/mol}$
