We can solve this equation using **Graham's law of effusion**:

#(r_1)/(r_2) = sqrt((MM_2)/(MM_1))#

We can substitute gas #"X"# and #"oxygen"# in for 1 and 2, respectively, to get

#(r_"X")/(r_ ("O"_2)) = sqrt((MM_ ("O"_2))/(MM_"X"))#

Since #"O"_2# effuses at a rate twice that of gas #"X"#, the ratio of the rate of effusion of #"X"# to #"O"_2# is #0.5#:

#0.5 = sqrt((MM_ ("O"_2))/(MM_"X"))#

Since the molar mass of #"O"_2# is given as #32.00# #"g/mol"#, we then have

#0.5 = sqrt((32color(white)(l)"g/mol")/(MM_"X"))#

Using algebra to solve for #MM_"X"#:

#0.5 = sqrt(32color(white)(l)"g/mol")/(sqrt(MM_"X"))#

#sqrt(MM_"X") = sqrt(32color(white)(l)"g/mol")/0.5#

#MM_"X" = (sqrt(32color(white)(l)"g/mol")/0.5)^2 = color(red)(128# #color(red)("g/mol"#

The molar mass of the unknown gas is thus #128# #"g/mol"#

A gas close to this value is #sfcolor(blue)("iodine"#, with molar mass #126.90# #"g/mol"#.