For a particular lab experiment, you need 150 ml of a 2.50 m solution of ammonium nitrate. how many grams of nh4no3 will you need for the solution?

8.0 g

10.02 g

25.0 g

875.0 g

30.0 g

mass NH4NO3 = 30.0 g

Explanation:

∴ V = 150 mL

∴ C NH4NO3 = 2.50 M

⇒ mass NH4NO3 = ?

∴ mol NH4NO3 = (0.150 L)×(2.50 mol/L) = 0.375 mol NH4NO3

∴ Mw NH4NO3 = 80.043 g/mol

⇒ mass NH4NO3 = (0.375 mol)×(80.043 g/mol) = 30.016 g

some sources quote the lethal dose, for an average 170lbs opiate-naive individual is between 75 and 375 mg