According to the hess's law, the enthalpy of solution is: δhsoln = δhsolute + δhsolvent + δhmix. considering dissolving a small drop of nonpolar pentane oil in water which term(s) in the equation will experience the largest enthalpy change.

Here, we have to choose the effect of the enthalpy change on addition of pentane in water, as per the Hess's law.

The largest change of enthalpy will be found in the solute.  

The Hess's law of constant heat summation states that for a given chemical process the net heat change will be same irrespective of the steps involve in this.

Now, for addition of non polar pentane in water there basically no reaction will happen and it will behave as a heterogeneous mixture. However this addition will affect the δH(solute) and δH(mixture).

Due to the presence of non-polar molecule in highly polar solvent (H₂O) there will be no change of water molecules but the solute will shrink in a less volume, which will affect the van der waal's force/London Dispersion force of attraction between the molecules.

Both these attractions will increase the enthalpy of the solute. On the other hand the enthalpy of the mixture will also change.

The largest change of enthalpy will be found in the solute.  

i think is conclusions.i think so