Chemistry 10th Edition

n= $\frac{1.55g}{131.3gmol^{-1}}$ = 0.012 mol T= (20.+273) K = 293 K V= 560 mL = .560 L P = $\frac{nRT}{V}$ (using ideal gas law) = $\frac{0.012mol(0.0821\frac{L.atm}{mol.K})293K}{.560 L}$ $\approx$ 0.515 atm