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.
c₀(N₂O₅) = 0,015 mol/L.
c(N₂O₅) = 0,019 mol/ 1,8 L = 0,01055 mol/L.
k = 6,82·10⁻³ s⁻¹.
ln c(N₂O₅) = ln c₀(N₂O₅) - k·t.
t = (ln c₀(N₂O₅) - ln c(N₂O₅)) ÷ k.
t = 0,35 ÷ 0,00682 s⁻¹.
t = 51 s = 0,86 min.