Answer is: the molarity of a solution is 0.276 M.
V(solution) = 855 mL ÷ 1000 mL/L.
V(solution) = 0.855 L.
m(NH₄NO₃) = 18.9 g; mass of ammonium nitrate.
M(NH₄NO₃) = 80.04 g/mol; molar mass of ammonium nitrate.
n(NH₄NO₃) = m(NH₄NO₃) ÷ M(NH₄NO₃).
n(NH₄NO₃) = 18.9 g ÷ 80.04 g/mol.
n(NH₄NO₃) = 0.236 mol; amount of substance.
c(NH₄NO₃) = n(NH₄NO₃) ÷ V(solution).
c(NH₄NO₃) = 0.236 mol ÷ 0.855 L.
c(NH₄NO₃) = 0.276 mol/L; molarity of ammonium nitrate.
