#"Concentration"# #=# #"Moles of solute"/"Volume of solution"#, and typically the units of concentration are #mol*L^-1#.
So if we want to access the moles of solute, we take the product:
#"Concentration"xx"Volume of solution"#.
Here, #0.300*mol*cancel(L^-1)xx0.250*cancelL=??mol#, and cancellation of units gives us an answer in #"moles"# as required. This is an example of dimensional analysis. It would be all too simple to multiply instead of divide or vice versa; we have all done it. If you do such problems dimensionally, this is an extra check on the solution to your problem.
