Answer
$51KJ/mol,-8.7C^{\circ}$
Work Step by Step
We know that
$\Delta H_{soln}=-\Delta H_{lattice}+\Delta H_{hydr}$
We plug in the known values to obtain:
$\Delta H_{soln}=(-599KJ/mol)+(-548KJ/mol)=51KJ/mol$
We find the heat gained by the solute
$q_{KClO_4}=(10.0gKClO_4)(\frac{1mol\space KClO_4}{138.5489g\space KClO_4})(\frac{51KJ}{1mol.KClO_4})=3.610KJ$
and mass of solution is
$m_{soln}=\frac{100.0mL\space sol(1.05g\space sol)}{1mL\space sol}$
$m_{soln}=105g\space soln$
Now $\Delta T=\frac{q_{soln}}{(m_{soln})(C)}$
We plug in the known values to obtain:
$\Delta T=\frac{-3.6810\times 10^3J}{(105g)(4.05\frac{J}{g.C^{\circ}})}$
$\Delta T=-8.7C^{\circ}$
