Let Q(t) represent the amount of salt after t minutes.

Then,

dQ/dt = Rate in - Rate out

The rate is is (x g/liter)*(2 liters/min) = 2x g/min.

The rate out is (Concentration)*(2 liters/min)

Now the concentration = Amount/Volume = Q/120

Thus, rate out is (Q/120 g/liter)(2 liters/min)=Q/60 g/min.

Thus, the differencial equation is,

dQ/dt = 2x - Q/60

Thus,

dQ/dt + Q/60 = 2x

With initial conditions Q(0)=0.

This is a first order linear differencial equation, it solves easily. I am sure you can do it from there.