first, we should get K, and we can get it from this equation:
K= ㏑(2)/t1/2
when we have t1/2 (the half-life time )= 90 min so by substitution:
K= ㏑(2) / 90 min
=0.0077
Then we can get C ( the amount remains at time t) from this equation:
C= Ae^(-Kt)
we have A(amount at t=zero) = 2 g / mL
& K = 0.0077 & t = 2.5 h, convert it to min 2.5*60 = 150 m
by substitution:
C= 2* e^(-0.0077*150)
= 2 * 0.315 = 0.63 g/mL