If the process is asumed continuum you can calculate the average flow of methanol as (8,2 [g]) / (8*60 [min]*32 [g/mol])=5,338 *10^-4 [mol /min], the flow of nitrogen is (0,060 [l/min]*1 [atm])/(0,0821 [atm l /mol K]*(27+273,15)[K])=2,4348 *10^-3 [mol]
Now you can calculate the molar fraction leaving:
Y (CH3OH)= 5,338*10^-4 / ( 5,338*10^-4+2,4348*10^-3)=0,1798 [-]
Maybe I wrong the numbers, but keep the idea.