Since the ΔH is associated with a chemical equation, there is no need to write the unit as kJ/mol
2H2O2 ---> 2H2O + O2 ΔH = -196 kJ
In fact, the ΔH value is actually associated with 2H2O2 and if the term mol was to be included, the ΔH would have to be divided by 2.
The solution to the problem:
(-98 kJ/mol) (0.01 mol) = -0.98 kJ