Question

8.50g of NH3 is present in 250ml volume . Its active mass is

Answer 1

active mass = (wt/Mw) x (1000/V)
wt = weight of solute
Mw = molecular weight of solute
V = volume of solution in ml 

so 
active mass = (8.5/17) x (1000/250)
active mass = 0.5 x 4 = 2 M

regards

Answer 2

Active mass = gram moles per liter of solution
          = 8.50/17 *1/0.250 = 2 moles/L or 2 M

Molar mass of Ammonia = 14+3=17