How many minutes will it take for the quantity of n2o5 to drop to 1.9*10-2 mol?

Missing question: The first-order rate constant for the decomposition of N2O5, 2N2O5 (g) →4NO2 (g) + O2 (g) at 70° C is 6.82*10-3 s-1. Suppose we start with 2.70*10-2 mol of N2O5 (g) in a volume of 1.8 L.

c₀ (N₂O₅) = 0,027 mol : 1,8 L.

c ₀ (N₂O₅) = 0,015 mol/L.

c (N ₂O₅) = 0,019 mol / 1,8 L = 0,01055 mol/L.

k = 6,82·10 ⁻³ s⁻¹.

ln c (N ₂O₅) = ln c₀ (N₂O₅) - k·t.

t = (ln c ₀ (N₂O₅) - ln c (N₂O₅)) : k.

t = 0,35 : 0,00682 s ⁻¹.

t = 51 s = 0,86 min.