You will actually get a mixture of different products, with the product with the highest yield being p-chlorotoluene, or 1-chloro-4-methylbenzene. The other products resulting from the reaction include o-chlorotoluene and m-chlorotoluene, with a significantly higher amount of the former being produced*. In fact, there would also be products with more than one methyl substituent being produced in various proportions depending on the precise conditions being used.
Yes, the reaction will proceed by Friedel-Crafts alkylation: Firstly, the Lewis acid catalyst abstracts the chloride ion from methyl chloride, giving a reactive, electrophilic methenium ion. Then, electrons from the $\pi$ system of the benzene ring of chlorobenzene then attack the electrophilic ion. Finally, a proton is lost to restore aromaticity of the ring, giving the product.
*Refer to the following post to understand why the chlorine atom as a substituent would be ortho/para-directing: Why are halogens ortho para directing even though deactivating?