This is the Finkelstein reaction.Sodium iodide (NaI) reacts with methyl chloride ($\mathrm{CH_3Cl}$).Iodide ion $(\mathrm{I}^-)$ replaces chloride $(\mathrm{Cl}^-)$.Dry acetone is used because NaCl is insoluble in acetone, so it precipitates out, driving the reaction forward.Products:A = $\mathrm{CH_3I}$ (methyl iodide)B = NaCl (precipitate)